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For the matrix A= $$ \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{pmatrix} $$

how would I orthogonally diagonalize it. I can see quite plainly that it is a symmetric matrix. Does this help at all?

Thanks

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  • $\begingroup$ It helps in so far as you know that it can be diagonlized (with orthogonal basis). How do you diagonalize a matrix? $\endgroup$ – Doug M Mar 29 '17 at 1:32
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You can go through the usual process of solving the characteristic polynomial and computing various matrix kernels to find the eigenvalues and eigenvectors of this matrix, but the matrix is simple enough that one can quickly find them by inspection.

Recall that the product of a matrix and a (column) vector is a linear combination of the columns of the matrix, with coefficients given by the components of the vector. We see that the first column consists of all zeros, so it’s clear that $(1,0,0)^T$ is an eigenvector with eigenvalue $0$. Next, adding the last two columns produces $(0,1,1)^T$, which is exactly the vector that corresponds to adding the last two columns, so this vector is also an eigenvector with eigenvalue $1$. The trace of a square matrix is equal to the sum of its eigenvalues, so we know immediately that the third eigenvalue is $-1$. By subtracting the last column from the second, it’s not hard to see that $(0,1,-1)^T$ is an eigenvector of $-1$, but since this is the last eigenvalue, we have another easy way to find the last eigenspace: the eigenspaces of a real symmetric matrix that correspond to different eigenvalues are orthogonal, so our third eigenvector will be orthogonal to the first two. We can compute the orthogonal complement of the span of those two vectors in the usual way, or take advantage of working in $\mathbb R^3$ and simply compute their cross product, which is either $(0,-1,1)^T$ or $(0,1,-1)^T$ depending on the order in which you multiply them.

Now you have an orthogonal set of eigenvectors which form a basis of $\mathbb R^3$, so all that’s left is to normalize them.

Another way to find an orthogonal set of eigenvectors for this matrix is to use the fact that they are the principal axes of the related quadratic form. In this case, we have $$\pmatrix{x&y&z}\pmatrix{0&0&0\\0&0&1\\0&1&0}\pmatrix{x\\y\\z}=2yz.$$ This is a family of hyperbolic cylinders that parallel the $x$-axis, so we know that one of the principal axes is the $x$-axis, giving us the eigenvector $(1,0,0)^T$. The intersection of this cylinder with the $y$-$z$ plane is the hyperbola $yz=\text{const}$, so the other two principal axes are the axes of this hyperbola, $(0,1,1)^T$ and $(0,1,-1)^T$. With these vectors in hand, it’s a simple matter to find the corresponding eigenvalues if needed. As before, all that’s left to do is to normalize them to produce an orthonormal basis.

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Hint: (Orthogonal Diagonalization) Let $A$ be a $n × n$ matrix.

$1.$Solve the characteristic equation $\det (A − λIn)=0$ for $λ$ with multiplicity to find out eigenvalues.

$2.$ For each eigenvalue $λ$, find a basis of the eigenspace $Eλ = \ker (A − λIn)$.

$3$. Use Gram-Schmidt process to get an orthonormal basis from a basis of $(2)$ for each eigenspace $Eλ = \ker (A − λIn)$.

Now create a matrix $S$ with the vectors of orthonormal basis and this will work for you.

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