Dimension of Eigenspace?

Question

Given a matrix A

$ A = \begin{bmatrix} 5 & 4 & -1 \\ 4 & 5 & -1 \\ -4 & -4 & 2 \\ \end{bmatrix} $

I have to find out if A is diagonalizable or not. Also I have to write down the eigen spaces and their dimension.

For eigenvalue, $ \lambda = 1 $ , I found the following equation:

$ x_1 + x_2 - \frac{x_3}{4} = 0 $

Here, I have two free variables. $ x_2 $ and $ x_3 $. I'm not sure but I think the the number of free variables corresponds to the dimension of eigenspace and setting once $ x_2 = 0 $ and then $ x_3 = 0 $ will compute the eigenspace.

Any detailed explanation would be appreciated.

You're right: a single linear relation between $n$ variables makes a subspace of codimension $1$, i.e. of dimension $n-1$ — Bernard, Feb 13 '18 at 00:57

score 1 · Accepted Answer · answered Feb 13 '18 at 01:30

1

Set $x_3 = 0$ to get an eigenvector $v_1 = (1, -1, 0)^T$, then set $x_2 = 0$ to get another eigenvector $v_2 = (1, 0 , 4)^T$. To find the third eigenvalue $t$, write the characteristic polynomial as $$\det(\lambda-AI) = (\lambda-1)^2 (\lambda - t).$$ Substitute $\lambda = 0$ in the above equation to see that $-\det A = -t$, so \begin{align} t =& \det A = (5)(5)(2)+(4)(-1)(-4)+(-1)(4)(-4) \\ &- (-4)(5)(-1) - (-4)(-1)(5) - (2)(4)(4) \\ =& 50 + 16 + 16 - 20 - 20 - 32 = 10 \end{align}

$$(A-10I) \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix} -5 & 4 & -1 \\ 4 & -5 & -1 \\ -4 & -4 & -8 \\ \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = 0, \tag{*} \label1$$

$v_3 = (x_1,x_2,x_3) = (1,1,-1)$ solves \eqref{1}, so it's a eigenvector corresponding to eigenvalue $10$.

The eigenspace $E_1 = \mathrm{span}\{v_1,v_2\}$ corresponding to $\lambda = 1$ has dimension $2$; the eigenspace $E_{10} = \mathrm{span}\{v_3\}$ corresponding to $\lambda = 10$ has dimension $1$.

answered Feb 13 '18 at 01:30

GNUSupporter 8964民主女神地下教會

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How can I find out if the matrix is diagonalizable? – Robur_131 Feb 13 '18 at 02:16
A matrix is diagonalizable iff its eigenvectors in $\Bbb F^n$ form a basis for $\Bbb F^n$. – GNUSupporter 8964民主女神地下教會 Feb 13 '18 at 02:20
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So does the determinant of the eigen-vector matrix being non-zero implies A is diagonalizable? – Robur_131 Feb 13 '18 at 07:32
@Robur_131 Not necessarily. In fact, my prof says in 90% of the time, a square matrix is not diagonalizable. https://math.stackexchange.com/q/2207078/290189 – GNUSupporter 8964民主女神地下教會 Feb 13 '18 at 07:34
Do the eigen-vectors need to be linearly independent in order to form a basis for $ \mathbb{F}^n $ ? If not, how can I check? – Robur_131 Feb 13 '18 at 08:36
@Robur_131 sorry I mean on R, 90% of square matrices aren't diagonalizable. To form a basis, you always need linearly independent vectors, since the goal is to represent the subspace using a few representatives. – GNUSupporter 8964民主女神地下教會 Feb 13 '18 at 11:34

score 0 · Answer 2 · answered Feb 13 '18 at 01:45

Ummm If you can think of only one specific eigenvector for eigenvalue $1,$ with actual numbers, that will be good enough to start with. Call it $(u,v,w).$ It has a dot product of zero with $(4,4,-1.)$ We would like a second one. So, take second eigenvector $(4,4,-1) \times (u,v,w)$ using traditional cross product.

Dimension of Eigenspace?

2 Answers2