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3a of this released exam asks (paraphrased):

Diagonalize the matrix $$A = \begin{bmatrix} 0 & 4 & 0 \\ 4 & 0 & 4 \\ 0 & 4 & 0 \end{bmatrix}$$ by an orthogonal matrix.

I tried using the standard calculate the eigenvalues $\lambda_{i} = 0, 2(1 \pm \sqrt{2})$ and getting the appropriate eigenvector matrix for:

$$\begin{bmatrix} 1 & -1 & 1 \\ -\sqrt{2} & 0 & \sqrt{2} \\ 1 & 1 & 1 \end{bmatrix}$$

However if you multiply any permutation by the transpose, you get:

$$\begin{bmatrix} 3 & 0 & 1 \\ 0 & 4 & 0 \\ 1 & 0 & 3 \end{bmatrix}$$

instead of $I_n$. I'm stuck. Is there an easier way to compute an orthogonal matrix that diagonalizes $A$?

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  • $\begingroup$ The fastest surest way is to carry on Gram-Schmidt in your basis of eigenvectors and get thus an orthonormal basis from which construct a matrix $\;P\;$ for which $\;PAP^{-1}\;$ will be diagonal. $\endgroup$ – DonAntonio May 3 '16 at 7:29
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Your matrix is not orthogonal. Remember an orthogonal matrix has columns which are perpendicular and unit length. So your first column should be $$\frac12\pmatrix{1\cr-\sqrt2\cr1\cr}$$ and the others should be adjusted similarly.

Comment. An orthogonal matrix is one with orthonormal columns. The terminology is, frankly, very confusing, but it's been around a long time and it probably is not going to change :(

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  • $\begingroup$ I learned orthogonal as $A\cdot A^{T} = I_n$ and I wasn't really aware of the orthonormality that needed preservation. Normalizing each vector gives an identity matrix. Thanks! $\endgroup$ – Dair May 3 '16 at 7:39
  • $\begingroup$ It's a useful alternative characterisation: for real square matrices, $AA^T=I$ iff the columns of $A$ are orthonormal. $\endgroup$ – David May 3 '16 at 8:30
  • $\begingroup$ Note, I didn't check your eigenvalues and eigenvectors, from other people's answers it appears that they are wrong. But the principle is the same. $\endgroup$ – David May 3 '16 at 8:31
  • $\begingroup$ Yeah, sorry. I calculated the eigenvalues twice in my notebook and copied the wrong ones down. $\endgroup$ – Dair May 3 '16 at 8:32
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Quick note, your eigenvalues are actually $0, \pm4\sqrt{2}.$ The easiest way to compute the eigenbasis is to find the null space of $A-\lambda I$ for each eigenvalue $\lambda.$ The orthogonal bases of the necessary null spaces are: $$\text{null}(A-0\cdot I) = \text{null}(A) = \Bigg\langle\left({ \begin{array}{c} -\frac{\sqrt{2}}{2}\\0\\\frac{\sqrt{2}}{2} \end{array}}\right)\Bigg\rangle$$ $$\text{null}(A-4\sqrt{2}\cdot I) = \Bigg\langle\left({ \begin{array}{c} -\frac{1}{2}\\-\frac{\sqrt{2}}{2}\\-\frac{1}{2} \end{array}}\right)\Bigg\rangle$$ $$\text{null}(A+4\sqrt{2}\cdot I) = \Bigg\langle\left({ \begin{array}{c} \frac{1}{2}\\-\frac{\sqrt{2}}{2}\\\frac{1}{2} \end{array}}\right)\Bigg\rangle$$ These were easily found via Gaussian Elimination. Thus the matrix $$Q = \left({ \begin{array}{ccc} -\frac{\sqrt{2}}{2} & -\frac{1}{2} &\frac{1}{2}\\0 & -\frac{\sqrt{2}}{2} &-\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2} & -\frac{1}{2} & \frac{1}{2} \end{array}}\right)$$ orthogonally diagonalizes $A$ since $A = Q \Lambda Q^T$ where $\Lambda$ is the diagonal matrix with the eigenvalues on the diagonal.

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The matrix that diagonalizes $$ A=\left( \begin{array}{ccc} 0 & 4 & 0 \\ 4 & 0 & 4 \\ 0 & 4 & 0 \\ \end{array} \right) $$

is $$ P=\left( \begin{array}{ccc} -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ \frac{1}{2} & -\frac{1}{\sqrt{2}} & \frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{\sqrt{2}} & -\frac{1}{2} \\ \end{array} \right) $$ with the following formula: $$ P \cdot A \cdot P^T=D $$ and doing the calculation you obtain: $$ D=\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & -4 \sqrt{2} & 0 \\ 0 & 0 & 4 \sqrt{2} \\ \end{array} \right) $$

also note that $P$ is unitary: $$ P\cdot P^t={I}_{(3)} $$

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