Find all eigenvalues and eigenvectors of matrix

A=⎡⎣⎢11−2002−101⎤⎦⎥

are those ceiling and floor symbols? And what matrix? Is that a 1-entry matrix? Surely you can find the eigenstuff for a 1x1 matrix!

To find the eigenvalues and eigenvectors of matrix A, we need to solve the characteristic equation.

First, let's calculate the determinant of A minus λI, where I is the identity matrix and λ is the eigenvalue we are trying to find.

A - λI = ⎡⎣⎢11− λ−2002−10− λ1⎤⎦⎥

Next, we find the determinant of A - λI:

det(A - λI) = (11 - λ)(- λ) - (-2)(2 - λ)
= (11λ - λ^2) + 4 - 2λ
= -λ^2 + 9λ + 4

Setting the determinant equal to zero and solving for λ:

-λ^2 + 9λ + 4 = 0

Factoring this quadratic equation, we get:

(-λ - 1)(λ - 4) = 0

So the eigenvalues are λ = -1 and λ = 4.

To find the corresponding eigenvectors, we substitute each eigenvalue back into A - λI and solve for the eigenvectors.

For λ = -1:
A - (-1)I = ⎡⎣⎢11+1002−10−1⎤⎦⎥ = ⎡⎣⎢22 1−100−11⎤⎦⎥

Reducing this matrix to row-echelon form:
⎡⎣⎢22 1−100−11⎤⎦⎥ → ⎡⎣⎢1 0− 51 − 11/2⎤⎦⎥

From this, we can obtain the eigenvector:

x1 = 5x2

Therefore, the eigenvector corresponding to λ = -1 is v1 = ⎡⎢⎢⎣51⎤⎥⎥⎦ (where x2 is a free parameter).

For λ = 4:
A - 4I = ⎡⎣⎢11−4002−14⎤⎦⎥ = ⎡⎣⎢−3−4−400−1⎤⎦⎥

Reducing this matrix to row-echelon form:
⎡⎣⎢−3−4−400−1⎤⎦⎥ → ⎡⎣⎢1 4 0 1/3⎤⎦⎥

From this, we can obtain the eigenvector:

x1 = -4x2 - (1/3)x4

Therefore, the eigenvector corresponding to λ = 4 is v2 = ⎡⎢⎢⎣−41−31⎤⎥⎥⎦ (where x2 and x4 are free parameters).

To summarize:
Eigenvalue λ = -1 has eigenvector v1 = ⎡⎢⎢⎣51⎤⎥⎥⎦
Eigenvalue λ = 4 has eigenvector v2 = ⎡⎢⎢⎣−41−31⎤⎥⎥⎦

To find the eigenvalues and eigenvectors of a matrix, we need to solve the equation (A - λI)v = 0, where A is the matrix, λ is the eigenvalue, I is the identity matrix, and v is the eigenvector.

Let's find the eigenvalues:

First, let's calculate the determinant of (A - λI):

A - λI = ⎡⎣⎢11−λ00−10−1−λ1⎤⎦⎥

Determinant of (A - λI) = (1-λ)((-1)(-1-λ)) - (0)(0-(-1)) - (0)(1-(-1))
= (1-λ)(1+λ) - 0 - 0
= 1-λ+λ-λ^2 - 0
= -λ^2+2λ+1

Now, set the determinant equal to zero and solve for λ:

-λ^2+2λ+1 = 0

This is a quadratic equation, so we can factor it or use the quadratic formula to solve for λ:

Factorizing: (-λ+1)(λ+1) = 0

So, we have two eigenvalues: λ = 1 and λ = -1.

Now, let's find the eigenvectors corresponding to each eigenvalue:

For λ = 1:

(A - λI)v = 0
(⎡⎣⎢11−2002−101⎤⎦⎥ - 1⎡⎣⎢1000100⎤⎦⎥)v = 0
⎡⎣⎢0012−101⎤⎦⎥v = 0

Reducing the matrix row echelon form:
⎡⎣⎢0100−12−101⎤⎦⎥v = 0

From the last row, we can see that x3 = x1 + 2x2.
So, let's choose x2 = 1, which gives x3 = 1 + 2(1) = 3.

Let's choose x1 = 0, then we have:
⎡⎣⎢0100013⎤⎦⎥v = 0

Let's choose x2 = 0, then we have:
⎡⎣⎢0100013⎤⎦⎥v = 0

So, one eigenvector is [0, 1, 3].

For λ = -1:

(A - λI)v = 0
(⎡⎣⎢11−2002−101⎤⎦⎥ - (-1)⎡⎣⎢1000100⎤⎦⎥)v = 0
⎡⎣⎢2012−100⎤⎦⎥v = 0

Reducing the matrix row echelon form:
⎡⎣⎢010−20020−100⎤⎦⎥v = 0

From the last row, we can see that x3 = -2x1 + 2x2.
Let's choose x1 = 1, which gives x3 = -2(1) + 2(1) = 0.

Let's choose x2 = 0, then we have:
⎡⎣⎢1000010⎤⎦⎥v = 0

So, another eigenvector is [1, 0, 0].

Therefore, the eigenvalues of the matrix A are λ = 1 and λ = -1, and the corresponding eigenvectors are [0, 1, 3] and [1, 0, 0], respectively.