Confirm that matrix A is similar to diagonal matrix

Question

I have to confirm that matrix A is similar to diagonal matrix. I have to find diagonal matrix D and rotation matrix P so it will be $D=P^-1AP.$

Matrix $A=\begin{bmatrix}1&0&0&2\\0&2&1&0\\ 0&2&1&0\\1&0&0&2\end{bmatrix}$.

My question is how do I find that P matrix and as D I would use matrix that looks something like $D=\begin{bmatrix}x&0&0&0\\0&x&0&0\\ 0&0&x&0\\0&0&0&x\end{bmatrix}$ and then find the x. Is my understanding correct or is it totally out of the way? (I'm not asking for solution just for hints and if I understand the problem correctly).

Answer 1

Find a basis of eigenvectors of $A$. Then the matrix whose columns are those eigenvectors is the matrix $P$ that you're after.

Answer 2

You have to find the eigenvalues $\lambda_1, \lambda_2, \lambda_3, \lambda_4 $ of the matrix $A$, that is, the roots of de polynomial $p(\lambda):= \mbox{ det}(A-\lambda I)$. Then you can take $D$ as $$D=\begin{pmatrix} \lambda_1 & 0 & 0 & 0 \\ 0 & \lambda_2 & 0 & 0 \\ 0 & 0 & \lambda_3 & 0 \\ 0 & 0 & 0 & \lambda_4 \end{pmatrix}.$$ For $P$ you have to find an orthonormal basis $v_1, v_2, v_3, v_4$ of $\mathbb{R}^4$ such that $$Av_i= \lambda_iv_i, \ \mbox{ for } i=1,2,3,4,$$

and you can take the column $i$ of $P$ as the vector $v_i$.

Answer 3

Your understanding is close to correct. The question is asking you to find a change of basis (or change of co-ordinates) such that the matrix $A$ is diagonal when considered in the new co-ordinate system. The matrix $P$ acts as a change of co-ordinate or change of basis matrix, and the matrix $D$ acts to scale the new basis vectors.

To find the correct co-ordinate system, the idea is to find a set of vectors $v_1,v_2,v_3,v_4$ such that $A$ scales these vectors without rotating them. This is equivalent to ensuring that the vector $Av_i$ is parallel to $v_i$. Written as an equality, we must impose $Av_i = \lambda_i v_i$. The vectors $v_i$ are exactly the eigenvectors of the matrix $A$ with corresponding eigenvalue $\lambda_i$. To find the eigenvalues $\lambda_i$, we must solve the polynominal system $\det(A-\lambda I)= 0$.

If the eigenvectors form a basis of $\mathbb{R}^4$, then we have found the change of co-ordinates we were looking for at the very beginning and $ P = [v_1, v_2, v_3, v_4]$. It should be fairly simple to see that we can scale the vectors $v_i$ to ensure that their length is 1. However, these vectors are not necessarily orthogonal, and $P$ isn't necessarily a rotation matrix (if $A$ is symmetric, then the eigenvectors are orthogonal).

Since the matrix $A$ scales the new co-ordinates by their respective eigenvalue $\lambda_i$, the elements of $D$ are precisely the eigenvalues we found earlier. Here, your guess that $$ D = \left[ \begin{array}{cccc} x & 0 & 0 & 0 \\ 0 & x & 0 & 0 \\ 0 & 0 & x & 0 \\ 0 & 0 & 0 & x \\ \end{array} \right], $$ isn't quite right, as the diagonal elements don't have to be the same value $x$. Rather, the diagonal elements are the eigenvalues, $\lambda_i$.

Answer 4

As you can observe that elements that you want on your diagonal are $ 0, 0, 3, 3$ which are infact the eigenvalues. So that gives your characteristic polynomial as $c_A(x)= x^2(x-3)^2$. If you want your matrix to be a strictly disgonal matrix then you have to verify that the minimal polynomial is $m_A(x)=x(x-3)$. The matrix $P$ you are looking for is the matrix of eigenvectors corresponding to your eigenvalues $0,3$. One of the columns of your $P$ wpuld be $$\begin{bmatrix}1 \\1\\1\\1\end{bmatrix}$$

Answer 5

Actually you don't need any numerical calculation to prove that $A$ is diagonalisable. Clearly $A$ is similar to $$ A([1,4,3,2],[1,4,3,2])=\left[\begin{array}{cc|cc}1&2\\ 1&2\\ \hline&&1&2\\ &&1&2\end{array}\right]. $$ So, it suffices to prove that $X=\pmatrix{1&2\\ 1&2}$ is diagonalisable.

Since $X$ has rank one, one of its eigenvalue is zero. However, as $X$ is entrywise positive, $X^2\ne0$. Hence $X$ isn't nilpotent and its other eigenvalue must be nonzero. Therefore the two eigenvalues of $X$ are distinct and $X$ is diagonalisable.