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Consider the matrix

$$A=\left(\begin{array}{cccc} 1 &1 &-3 &0 \\ -1 &-1 &1 &-2 \\ -1 &-1 &-1 &-2 \\ 1 &1 &2 &3 \\ \end{array}\right)$$

The characteristic polynomial of $A$ is $\chi _A(x)=x^2(x-1)^2.$ Is $A$ diagonalizable over $\mathbb{C}?$ Explain.

I'm having a hard time understanding diagonalization. I believe that the characteristic polynomial yields 2 distinct eigenvalues $(0$ and $1)$. From what I understand, if there were $4$ distinct eigenvalues, you would know for sure that it's diagonalizable. However, since there are only $2$, where do you go from here to decide if it's digonalizable or not?

shprecure
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  • You found that the algebraic multiplicity of the eigenvalues $0$ and $1$ are both two. Check now to see if the geometric multiplicity of the eigenvalues are also two or not. Do this by computing the dimension of the kernel of $(A-\lambda I)$. If algebraic multiplicity equals geometric multiplicity for every eigenvalue, then it is indeed diagonalizable. (note: geometric multiplicity is always at least one for an eigenvalue and is bounded above by algebraic multiplicity. This is why having distinct eigenvalues guarantees diagonalizability) – JMoravitz Nov 19 '15 at 20:43
  • Okay, so the dimension of the kernel is $1$ for the eigenvalue $0$, and $2$ for the eigenvalue $1$. This means that the algebraic multiplicity isn't equal to the geometric multiplicity of each eigenvalue, and therefore the matrix isn't diagonalizable. Is that a step in the correct direction? – shprecure Nov 19 '15 at 20:51

1 Answers1

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Let $\lambda$ be an eigenvalue. You should check the eigenspaces, i.e the kernel of the matrix $(A-\lambda I_4)$. If a vector $v$ is in the kernel $\text{Ker}(A-\lambda I_4)$ of the matrix, then

$$(A-\lambda I_4)(v)=0 \iff A(v)-\lambda v=0 \iff A(v)=\lambda v$$

Vectors like $v$ are called eigenvectors. If you can find two linearly independent vectors $v_1,v_2 \in \text{Ker}(A-\lambda_1)$ for the first eigenvalue, another two linearly independent vectors $w_1,w_2 \in \text{Ker}(A-\lambda_2)$ for the second value you can form a basis $\mathcal{V}=(v_1,v_2,w_1,w_2)$. Since the columns of a matrix are the images of your basis' vectors, the matrix will be diagonal. It should look like this:

$$\begin{pmatrix}\lambda_1 & 0 & 0 & 0\\ 0 & \lambda_1 & 0 & 0 \\ 0 & 0 & \lambda_2 & 0 \\ 0 & 0 & 0 & \lambda_2 \end{pmatrix} = \begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$

In a non-algebraically closed field (i.e. $\mathbb{R}$) you should also check if all the roots of the characteristic polynomial lie in the field. This is an intrinsic characteristic of the vector space we are working on. Here, because we are working on $\mathbb C$, we have no problems.

If the geometric multeplicity $M_g (\lambda):= \text{dim(Ker }(A-\lambda)$ i.e. the dimension of the kernel of the matrix $A-\lambda I_4$ is strictly smaller than the algebraic multeplicity $M_a (\lambda)$ of $\lambda$ as a root of the characteristic polynomial, the matrix will not be diagonalizable. We can verify through algebra that $M_g (\lambda)\leq M_a (\lambda)$ is always true, but the matrix is diagonalizable if and only if equality holds.

What are the dimensions of the two kernels? Can you find these four vectors?

Lonidard
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  • I found that the dimension of the $ker(A-0I_4)$ is $1$ and the corresponding vector is $(-1,1,0,0)$ and that the dimension of the $ker(A-1I_4)$ is $2$ and the corresponding vectors are $(-5,3,1,0)$ and $(-2,0,0,1)$. So, according to these results, the geometric multiplicity is smaller than the algebraic multiplicity, and the matrix is not diagonalizable. Is that right? – shprecure Nov 19 '15 at 21:19
  • @shprecure Yes, perfect. Those are the correct eigenvectors. Good job! – Lonidard Nov 19 '15 at 21:28