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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $D=\operatorname{diag}(-2,1,1)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.