Real 2×2 Eigenvalue and Nonzero Eigenvector Foundations

No

Mv=[0, 0] and λv=[0, 0]. Both equality and v≠0 are required, giving No. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

Yes

Mv=[0, 3] and λv=[0, 3]. Both equality and v≠0 are required, giving Yes. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

No

Mv=[-10, 4] and λv=[-10, -10]. Both equality and v≠0 are required, giving No. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

No

Mv=[-9, -5] and λv=[-8, 8]. Both equality and v≠0 are required, giving No. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

Yes

Mv=[5, 5] and λv=[5, 5]. Both equality and v≠0 are required, giving Yes. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

Yes

Mv=[5, -5] and λv=[5, -5]. Both equality and v≠0 are required, giving Yes. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

No

Mv=[-6, 8] and λv=[-6, -6]. Both equality and v≠0 are required, giving No. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

No

Mv=[-5, -9] and λv=[-4, 4]. Both equality and v≠0 are required, giving No. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

Yes

Mv=[-4, 0] and λv=[-4, 0]. Both equality and v≠0 are required, giving Yes. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

Yes

Mv=[0, 11] and λv=[0, 11]. Both equality and v≠0 are required, giving Yes. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

No

Mv=[-2, 12] and λv=[-2, -2]. Both equality and v≠0 are required, giving No. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

No

Mv=[-1, -13] and λv=[0, 0]. Both equality and v≠0 are required, giving No. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

Yes

Mv=[0, 0] and λv=[0, 0]. Both equality and v≠0 are required, giving Yes. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

Yes

Mv=[0, 15] and λv=[0, 15]. Both equality and v≠0 are required, giving Yes. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

No

Mv=[2, 16] and λv=[2, 2]. Both equality and v≠0 are required, giving No. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

No

Mv=[3, -17] and λv=[4, -4]. Both equality and v≠0 are required, giving No. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

Yes

Mv=[4, 0] and λv=[4, 0]. Both equality and v≠0 are required, giving Yes. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

Yes

Mv=[0, 19] and λv=[0, 19]. Both equality and v≠0 are required, giving Yes. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

No

Mv=[6, 20] and λv=[6, 6]. Both equality and v≠0 are required, giving No. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

No

Mv=[7, -21] and λv=[8, -8]. Both equality and v≠0 are required, giving No. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

Yes

Mv=[8, 0] and λv=[8, 0]. Both equality and v≠0 are required, giving Yes. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

Yes

Mv=[0, 23] and λv=[0, 23]. Both equality and v≠0 are required, giving Yes. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

No

Mv=[10, 24] and λv=[10, 10]. Both equality and v≠0 are required, giving No. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

No

Mv=[11, -25] and λv=[12, -12]. Both equality and v≠0 are required, giving No. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

Yes

Mv=[12, 0] and λv=[12, 0]. Both equality and v≠0 are required, giving Yes. The zero vector is not an eigenvector, even though the equality then holds; zero or negative eigenvalues are not automatically excluded.

{0}

The two diagonal values give {0}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{-2}

The two diagonal values give {-2}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{-10, 3}

The two diagonal values give {-10, 3}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{-9, 4}

The two diagonal values give {-9, 4}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{-8, 5}

The two diagonal values give {-8, 5}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{-7, 6}

The two diagonal values give {-7, 6}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{-6, 7}

The two diagonal values give {-6, 7}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{-5, 8}

The two diagonal values give {-5, 8}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{-4, 9}

The two diagonal values give {-4, 9}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{-3, 10}

The two diagonal values give {-3, 10}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{-2, 11}

The two diagonal values give {-2, 11}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{-1, 12}

The two diagonal values give {-1, 12}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{0, 13}

The two diagonal values give {0, 13}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{1, 14}

The two diagonal values give {1, 14}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{2, 15}

The two diagonal values give {2, 15}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{3, 16}

The two diagonal values give {3, 16}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{4, 17}

The two diagonal values give {4, 17}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{5, 18}

The two diagonal values give {5, 18}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{6, 19}

The two diagonal values give {6, 19}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{7, 20}

The two diagonal values give {7, 20}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{8, 21}

The two diagonal values give {8, 21}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{9, 22}

The two diagonal values give {9, 22}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{10, 23}

The two diagonal values give {10, 23}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{11, 24}

The two diagonal values give {11, 24}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

{12, 25}

The two diagonal values give {12, 25}. Equal values are listed once; this is a set of distinct eigenvalues, not an algebraic-multiplicity count.

Yes

The residual (M−λI)v is [0, 0], giving Yes. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

Yes

The residual (M−λI)v is [0, 0], giving Yes. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

No

The residual (M−λI)v is [0, 14], giving No. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

No

The residual (M−λI)v is [-1, -13], giving No. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

Yes

The residual (M−λI)v is [0, 0], giving Yes. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

Yes

The residual (M−λI)v is [0, 0], giving Yes. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

No

The residual (M−λI)v is [0, 14], giving No. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

No

The residual (M−λI)v is [-1, -13], giving No. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

Yes

The residual (M−λI)v is [0, 0], giving Yes. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

Yes

The residual (M−λI)v is [0, 0], giving Yes. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

No

The residual (M−λI)v is [0, 14], giving No. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

No

The residual (M−λI)v is [-1, -13], giving No. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

Yes

The residual (M−λI)v is [0, 0], giving Yes. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

Yes

The residual (M−λI)v is [0, 0], giving Yes. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

No

The residual (M−λI)v is [0, 14], giving No. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

No

The residual (M−λI)v is [-1, -13], giving No. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

Yes

The residual (M−λI)v is [0, 0], giving Yes. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

Yes

The residual (M−λI)v is [0, 0], giving Yes. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

No

The residual (M−λI)v is [0, 14], giving No. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

No

The residual (M−λI)v is [-1, -13], giving No. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

Yes

The residual (M−λI)v is [0, 0], giving Yes. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

Yes

The residual (M−λI)v is [0, 0], giving Yes. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

No

The residual (M−λI)v is [0, 14], giving No. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

No

The residual (M−λI)v is [-1, -13], giving No. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

Yes

The residual (M−λI)v is [0, 0], giving Yes. Every such kernel contains zero, and zero-vector membership does not establish that λ is an eigenvalue.

1

The characteristic polynomial coefficients are (1,0,0), with discriminant Δ=0. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 1. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

0

The characteristic polynomial coefficients are (1,-2,3), with discriminant Δ=-8. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 0. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

1

The characteristic polynomial coefficients are (1,-2,1), with discriminant Δ=0. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 1. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

2

The characteristic polynomial coefficients are (1,0,-2), with discriminant Δ=8. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 2. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

0

The characteristic polynomial coefficients are (1,-4,7), with discriminant Δ=-12. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 0. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

1

The characteristic polynomial coefficients are (1,-4,4), with discriminant Δ=0. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 1. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

2

The characteristic polynomial coefficients are (1,0,-5), with discriminant Δ=20. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 2. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

0

The characteristic polynomial coefficients are (1,-6,13), with discriminant Δ=-16. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 0. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

1

The characteristic polynomial coefficients are (1,-6,9), with discriminant Δ=0. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 1. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

2

The characteristic polynomial coefficients are (1,0,-10), with discriminant Δ=40. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 2. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

0

The characteristic polynomial coefficients are (1,-8,21), with discriminant Δ=-20. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 0. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

1

The characteristic polynomial coefficients are (1,-8,16), with discriminant Δ=0. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 1. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

2

The characteristic polynomial coefficients are (1,0,-17), with discriminant Δ=68. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 2. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

0

The characteristic polynomial coefficients are (1,-10,31), with discriminant Δ=-24. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 0. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

1

The characteristic polynomial coefficients are (1,-10,25), with discriminant Δ=0. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 1. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

2

The characteristic polynomial coefficients are (1,0,-26), with discriminant Δ=104. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 2. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

0

The characteristic polynomial coefficients are (1,-12,43), with discriminant Δ=-28. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 0. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

1

The characteristic polynomial coefficients are (1,-12,36), with discriminant Δ=0. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 1. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

2

The characteristic polynomial coefficients are (1,0,-37), with discriminant Δ=148. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 2. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

0

The characteristic polynomial coefficients are (1,-14,57), with discriminant Δ=-32. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 0. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

1

The characteristic polynomial coefficients are (1,-14,49), with discriminant Δ=0. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 1. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

2

The characteristic polynomial coefficients are (1,0,-50), with discriminant Δ=200. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 2. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

0

The characteristic polynomial coefficients are (1,-16,73), with discriminant Δ=-36. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 0. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

1

The characteristic polynomial coefficients are (1,-16,64), with discriminant Δ=0. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 1. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.

2

The characteristic polynomial coefficients are (1,0,-65), with discriminant Δ=260. Negative, zero, and positive Δ give zero, one, and two distinct real roots, so the answer is 2. A positive nonsquare Δ still gives two real roots; no diagonalizability, complex-spectrum, or Jordan conclusion follows here.