For every matrix $A$ we have $$AV=V\Lambda\rightarrow A=V\Lambda V^{-1}$$ where $V$ is the matrix of eigenvectors and $\Lambda$ is the diagonal matrix including eigenvalues.

When we have a symmetric matrix, one can write $$A^T=({V^{-1}})^T\Lambda V^T\rightarrow A=({V^{-1}})^T\Lambda V^T$$ where symmetry of $A$ results in $A^T=A$, and $\Lambda^T=\Lambda$ because it is diagonal.

Comparing these two yields $$V^{-1}=V^T\rightarrow V^TV=I$$ which is the definition of the orthogonal matrix. That means not only eigenvectors of a symmetric matrix are orthogonal but also they are orthonormal, i.e., they are unit vector with length 1 and orthogonal. Is this correct for all symmetric matrices?

  • $\begingroup$ If you have an orthogonal basis of eigenvectors, you can always turn it into an orthonormal basis of eigenvectors $\endgroup$
    – B. Mehta
    May 9, 2018 at 14:47
  • $\begingroup$ You made a huge jump when you “compare the two”. How did you arrive at $V^{-1} = V^T$? $\endgroup$
    – James Yang
    May 9, 2018 at 14:53

1 Answer 1


Comparing these two yields $$V^{-1}=V^T\rightarrow V^TV=I$$ which is the definition of the orthogonal matrix.

That step is wrong. I mean, it's not a bad guess to make. But suppose for a moment that you've found a $V$ such that $V^T = V^{-1}$.

Now write $W = 2V$. Then we'll have $$ A = W \Lambda W^{-1} $$ because the factors of $2$ and $\frac{1}{2}$ will cancel.

But we'll also have $$ A = (W^T)^{-1} \Lambda W^T $$ by the reasoning you presented. If the quoted remark above were valid, you would conclude that $W$ was ALSO orthonormal, which is impossible, because it's twice an orthonormal matrix.


It's true, however, that by taking the matrix $V$ of eigenvectors, you can produce an orthonormal matrix of eigenvectors: you just divide each column by its norm.

So your conclusion, that a symmetric matrix can be written as $U^t \Lambda U$, where $U$'s columns are an orthonormal basis of eigenvectors, is correct, but your method of reaching that conclusion was flawed.

  • $\begingroup$ I like you counter example. $\endgroup$
    – user494522
    May 9, 2018 at 15:18
  • $\begingroup$ @ John Is the only way to change orthogonal basis is to normalize the vectors of $A$ to get orthonormal basis? $\endgroup$
    – user494522
    Sep 26, 2018 at 16:27
  • $\begingroup$ You can start with any basis for your space and apply the Gram-Schmidt process to get a sequence of orthonormal vectors, which can then be used as the columns of an orthogonal matrix. The basis you start with need not have anything to do with $A$. Of course, when you do so, the resulting orthogonal matrix will be of no use in diagonalizing $A$ in general. $\endgroup$ Sep 26, 2018 at 16:32
  • $\begingroup$ Suppose we have such an orthogonal eigendecomposition $A=U^t\Lambda U$, and then normalise $U$, calling it $\overline{U}$. How can I see that $A=\overline{U}^t\Lambda \overline{U}$ again? It feels counter-intuitive that the transpose should act to cancel out the length factors, rather I feel like they would create a new matrix $\overline{A}$.. I apologise if this is confused $\endgroup$ Mar 6, 2019 at 8:53
  • 1
    $\begingroup$ @plebmatician: You're right -- they don't. Consider the case $U = U^t = 2I$, and $\Lambda = I$. Then $A = 4I$. But when we normalize to $\bar{U} = I$, we get $\bar{U}^t \Lambda \bar{U} = I$ instead of $4I$ -- to get the same answer, we need to alter $\Lambda$ (basically multiplying each diag. entry of $\Lambda$ by the squared length of the corresponding evec in $U$). In this example, we'd multiply each entry by $4$, to get $A = I (4I) I$, which is still correct. $\endgroup$ Mar 6, 2019 at 12:06

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