Let us assume that $$A=\left[\begin{array}{rrr}2&-1&-1\\-1&2&-1\\-1&-1&2\end{array}\right].$$

Then, eigenvalues of $A$ is $0, 3, 3$.

For $\lambda_1=0$, I could find its eigenvector as $\begin{bmatrix}1\\ 1\\ 1\end{bmatrix}$.

For $\lambda_1=3$, I could find its eigenvector as $\left[\begin{array}r0\\ 1\\ -1\end{array}\right]$ and $\left[\begin{array}r0\\ -1\\ 1\end{array}\right]$.

Thus, I set $S=\left[\begin{array}{rrr}1&0&0\\1&1&-1\\1&-1&1\end{array}\right]$ and $\Lambda=\begin{bmatrix}0&0&0\\0&3&0\\0&0&3\end{bmatrix}$.

The problem is $S\times \Lambda \times S^{-1}$ does not give $A$.

By matlab, $S$ is given as $\left[\begin{array}{rrr}0.5774&0.7634&0.2895\\0.5774&-0.6325&0.5164\\0.5774&-0.1310&-0.8059\end{array}\right]$

Using $S$ given by matlab, $S\times \Lambda \times S^{-1}$ gives $A$.

The textbook said the $i$th column of $S$ must be the eigenvector corresponding the $i$th eigenvalue of $A$.

I think this problem happens because two lambdas are same as $3$.

How can I find correct $S$?

  • 1
    $\begingroup$ Observe that the two eigenvectors you take for $\;\lambda=3\;$ are linearly dependent ... $\endgroup$ – DonAntonio Nov 26 '17 at 7:32


$$|xI-A|=\begin{vmatrix} x-2&1&1\\ 1&x-2&1\\ 1&1&x-2\end{vmatrix}=(x-2)^3+2-3(x-2)=x^3-6x^2+9x\implies$$

the eigenvalues are $\;0,\,3\;$, the last one of multiplicity two, and a basis for the corresponding eigenspaces is


Try to take it from here now.

  • $\begingroup$ Thank you, I got it. I should have chosen independent columns. May I ask one more thing? If I find $S$ such that $S^{-1}=S^{\text{T}}$ and $S\Lambda{S^{-1}}=A$, then $A=(S\sqrt{\Lambda})\times (S\sqrt{\Lambda})^\text{T}=K\times K^\text{T}$. Can I find all possible $T$? $\endgroup$ – Danny_Kim Nov 26 '17 at 7:54
  • $\begingroup$ @Danny_Kim You can find $\;S\;$ s.t. $\;S^{-1}=S^T\;$ ... I understand you haven't yet studied the spectral theorem in linear algebra? And what do you mean whether you can find all possible $\;T\;$ ?? What is $\;T\;$ for you in your comment? $\endgroup$ – DonAntonio Nov 26 '17 at 8:00
  • $\begingroup$ If I use Gram-Schumit formula, I can find $S$ such that $S^TS = I$, isn't it? I am ultinately wondering if we can find all possible matrix $T$ such that $A=T^TT$ $\endgroup$ – Danny_Kim Nov 26 '17 at 8:02
  • $\begingroup$ @Danny_Kim Yes, use GS on the eigenvectors' basis...and don't use the very same letter for both a matrix and its transpose. It is highly confusing...and observe that any orthogonal $\;S\;$ that diagonalizes $\;A\;$ , and there are infinitely many ones ones, will give you $\;A=SS^t\;$ . $\endgroup$ – DonAntonio Nov 26 '17 at 8:04
  • $\begingroup$ Ah, thank you for giving good comments. I just want to specify a set of matrices of $S$'s that satisfy $A=SS^\text{T}$ $\endgroup$ – Danny_Kim Nov 26 '17 at 8:08

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