Let A be a given matrix below. First, find the eigenvalues and their corresponding eigenspaces for the following matrices. Then, find an invertible matrix P and a diagonal matrix such that A = PDPâ’1.
(a) [ 3 2 2 3 ]
(b) [ 1 â 1 2 â 1 ]
(c) [1 2 3 0 2 3 0 0 3]
(d) [3 1 1 1 3 1 1 1 3]

It looks like given matrices are supposed to be

[tex]\begin{array}{ccccccc}\begin{bmatrix}3&2\\2&3\end{bmatrix} & & \begin{bmatrix}1&-1\\2&-1\end{bmatrix} & & \begin{bmatrix}1&2&3\\0&2&3\\0&0&3\end{bmatrix} & & \begin{bmatrix}3&1&1\\1&3&1\\1&1&3\end{bmatrix}\end{array}[/tex]

You can find the eigenvalues of matrix A by solving for λ in the equation det(A - λI) = 0, where I is the identity matrix. We also have the following facts about eigenvalues:

• tr(A) = trace of A = sum of diagonal entries = sum of eigenvalues

• det(A) = determinant of A = product of eigenvalues

(a) The eigenvalues are λ₁ = 1 and λ₂ = 5, since

[tex]\mathrm{tr}\begin{bmatrix}3&2\\2&3\end{bmatrix} = 3 + 3 = 6[/tex]

[tex]\det\begin{bmatrix}3&2\\2&3\end{bmatrix} = 3^2-2^2 = 5[/tex]

and

λ₁ + λ₂ = 6   ⇒   λ₁ λ₂ = λ₁ (6 - λ₁) = 5

⇒   6 λ₁ - λ₁² = 5

⇒   λ₁² - 6 λ₁ + 5 = 0

⇒   (λ₁ - 5) (λ₁ - 1) = 0

⇒   λ₁ = 5 or λ₁ = 1

To find the corresponding eigenvectors, we solve for the vector v in Av = λv, or equivalently (A - λI) v = 0.

• For λ = 1, we have

[tex]\begin{bmatrix}3-1&2\\2&3-1\end{bmatrix}v = \begin{bmatrix}2&2\\2&2\end{bmatrix}v = 0[/tex]

With v = (v₁, v₂)ᵀ, this equation tells us that

2 v₁ + 2 v₂ = 0

so that if we choose v₁ = -1, then v₂ = 1. So Av = v for the eigenvector v = (-1, 1)ᵀ.

• For λ = 5, we would end up with

[tex]\begin{bmatrix}-2&2\\2&-2\end{bmatrix}v = 0[/tex]

and this tells us

-2 v₁ + 2 v₂ = 0

and it follows that v = (1, 1)ᵀ.

Then the decomposition of A into PDP⁻¹ is obtained with

[tex]P = \begin{bmatrix}-1 & 1 \\ 1 & 1\end{bmatrix}[/tex]

[tex]D = \begin{bmatrix}1 & 0 \\ 0 & 5\end{bmatrix}[/tex]

where the n-th column of P is the eigenvector associated with the eigenvalue in the n-th row/column of D.

(b) Consult part (a) for specific details. You would find that the eigenvalues are i and -i, as in i = √(-1). The corresponding eigenvectors are (1 + i, 2)ᵀ and (1 - i, 2)ᵀ, so that A = PDP⁻¹ if

[tex]P = \begin{bmatrix}1+i & 1-i\\2&2\end{bmatrix}[/tex]

[tex]D = \begin{bmatrix}i&0\\0&i\end{bmatrix}[/tex]

(c) For a 3×3 matrix, I'm not aware of any shortcuts like above, so we proceed as usual:

[tex]\det(A-\lambda I) = \det\begin{bmatrix}1-\lambda & 2 & 3 \\ 0 & 2-\lambda & 3 \\ 0 & 0 & 3-\lambda\end{bmatrix} = 0[/tex]

Since A - λI is upper-triangular, the determinant is exactly the product the entries on the diagonal:

det(A - λI) = (1 - λ) (2 - λ) (3 - λ) = 0

and it follows that the eigenvalues are λ₁ = 1, λ₂ = 2, and λ₃ = 3. Now solve for v = (v₁, v₂, v₃)ᵀ such that (A - λI) v = 0.

• For λ = 1,

[tex]\begin{bmatrix}0&2&3\\0&1&3\\0&0&2\end{bmatrix}v = 0[/tex]

tells us we can freely choose v₁ = 1, while the other components must be v₂ = v₃ = 0. Then v = (1, 0, 0)ᵀ.

• For λ = 2,

[tex]\begin{bmatrix}-1&2&3\\0&0&3\\0&0&1\end{bmatrix}v = 0[/tex]

tells us we need to fix v₃ = 0. Then -v₁ + 2 v₂ = 0, so we can choose, say, v₂ = 1 and v₁ = 2. Then v = (2, 1, 0)ᵀ.

• For λ = 3,

[tex]\begin{bmatrix}-2&2&3\\0&-1&3\\0&0&0\end{bmatrix}v = 0[/tex]

tells us if we choose v₃ = 1, then it follows that v₂ = 3 and v₁ = 9/2. To make things neater, let's scale these components by a factor of 2, so that v = (9, 6, 2)ᵀ.

Then we have A = PDP⁻¹ for

[tex]P = \begin{bmatrix}1&2&9\\0&1&6\\0&0&2\end{bmatrix}[/tex]

[tex]D = \begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}[/tex]

(d) Consult part (c) for all the details. Or, we can observe that λ₁ = 2 is an eigenvalue, since subtracting 2I from A gives a matrix of only 1s and det(A - 2I) = 0. Then using the eigen-facts,

• tr(A) = 3 + 3 + 3 = 9 = 2 + λ₂ + λ₃   ⇒   λ₂ + λ₃ = 7

• det(A) = 20 = 2 λ₂ λ₃   ⇒   λ₂ λ₃ = 10

and we find λ₂ = 2 and λ₃ = 5.

I'll omit the details for finding the eigenvector associated with λ = 5; I ended up with v = (1, 1, 1)ᵀ.

• For λ = 2,

[tex]\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}v = 0[/tex]

tells us that if we fix v₃ = 0, then v₁ + v₂ = 0, so that we can pick v₁ = 1 and v₂ = -1. So v = (1, -1, 0)ᵀ.

• For the repeated eigenvalue λ = 2, we find the generalized eigenvector such that (A - 2I)² v = 0.

[tex]\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}^2 v = \begin{bmatrix}3&3&3\\3&3&3\\3&3&3\end{bmatrix}v = 0[/tex]

This time we fix v₂ = 0, so that 3 v₁ + 3 v₃ = 0, and we can pick v₁ = 1 and v₃ = -1. So v = (1, 0, -1)ᵀ.

Then A = PDP⁻¹ if

[tex]P = \begin{bmatrix}1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 0 & -1\end{bmatrix}[/tex]

[tex]D = \begin{bmatrix}5&0&0\\0&2&0\\0&2&2\end{bmatrix}[/tex]

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