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Find a matrix $P$ that orthogonally diagonalizes $I-uu^T$ if $u=\pmatrix{1\\-1\\1}$.

Let matrix $A=I-uu^T=\pmatrix{0&1&-1\\1&0&1\\-1&1&0}$.

I have followed the usual procedure and obtained matrix $P=\pmatrix{\frac{1}{\sqrt3}&-\frac{1}{\sqrt6}&\frac{1}{\sqrt2}\\-\frac{1}{\sqrt3}&\frac{1}{\sqrt6}&\frac{1}{\sqrt2}\\\frac{1}{\sqrt3}&\frac{2}{\sqrt6}&0}$.

In the midst of solving for $P^TAP$, I realized it got really tedious and long-winded. Is there is a better method to go about obtaining the diagonal matrix $D$ instead of computing it directly like so? I can't help but feel like I have missed something.

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  • $\begingroup$ Did you actually want the orthonormal eigenbasis (i.e., the diagonalizing matrix $P$), as you stated right at the top and in the question’s title, or did you simply way to find the diagonal matrix $D$? There are some short cuts for finding $P$ as well. $\endgroup$ – amd Apr 24 '17 at 23:35
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Note that the diagonal entries of $D$ are simply the eigenvalues of $A$. Once you confirm that the eigenvalues of $A$ are $-2,1,1$, you can immediately conclude that we have $$ D = \pmatrix{-2&0&0\\0&1&0\\0&0&1} $$ no computation of eigenvectors (or of $P$) required.

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  • $\begingroup$ Oh, I feel really silly now, thank you so much! Another question, if you don't mind: does the sequence of the column vector obtained for matrix $P$ matter? Sometimes I obtain a matrix $P$ with the same column entries but arranged differently from the solution. $\endgroup$ – iamaweed Apr 24 '17 at 22:07
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    $\begingroup$ The order of the columns in $P$ just needs to match up with the order of the eigenvalues in $D$. In this case, my version of $D$ would be the correct version since your first column is the eigenvector for $\lambda = -2$. $\endgroup$ – Ben Grossmann Apr 24 '17 at 22:11
  • $\begingroup$ Ah, I get it now. Thanks so much for your help! (: $\endgroup$ – iamaweed Apr 24 '17 at 22:12
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Assuming that you actually wanted the matrix $P$ as stated in the title and first sentence of your question, here are some short cuts.

Every eigenvector $v$ of a matrix $A$ is also an eigenvector of $I-A$: $$(I-A)v=v-A v=v-\lambda v=(1-\lambda)v.$$ We can also see from this that if $\lambda$ is an eigenvalue of $A$, then $1-\lambda$ is an eigenvalue of $I-A$.

The matrix $uu^T$ has rank one—every column is a scalar multiple of $u$— and its kernel consists of all vectors $v$ for which $u^Tv=0$, i.e., the orthogonal complement of $u$. Thus, an orthonormal basis of $u^\perp$ together with $u/\|u\|$ is an orthonormal basis of $\mathbb R^3$ that consists of eigenvector of $I-uu^T$, so you can take as $P$ the matrix with these vectors as its columns. The eigenvalues of $uu^T$, incidentally, are $0$ and $u^Tu$. The former is obvious because the nullity of $uu^T$ is $2$, while the latter is easily found by rearranging $(uu^T)u$.

For this problem, you can take another short cut since you’re working in $\mathbb R^3$: if you take any vector $v$ that’s orthogonal to $u$ and then compute $u\times v$, the three vectors form an orthogonal basis and all that’s left to do is to normalize them. You can find by inspection that $(1,1,0)^T$ is orthogonal to $u$, and $(1,-1,1)^T\times(1,1,0)^T=(-1,1,2)^T$, from which a quick computation produces $$P=\left[\begin{array}{rcr}\frac1{\sqrt3}&\frac1{\sqrt2}&-\frac1{\sqrt6}\\-\frac1{\sqrt3}&\frac1{\sqrt2}&\frac1{\sqrt6}\\\frac1{\sqrt3}&0&\frac2{\sqrt6}\end{array}\right].$$ The matrix $D=P^{-1}(I-uu^T)P$ has the corresponding eigenvalues along its main diagonal and $u^Tu=3$, so $D=\operatorname{diag}(-2,1,1)$.

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