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WLOG, we assume that

  • $\mathbf{A}$ is an $n$ by $n$ real and symmetric matrix,
  • $\lambda_1, \ldots, \lambda_n$ are eigenvalues, and
  • $\mathbf{s}_1, \ldots, \mathbf{s}_n$ are eigenvectors of size $n$ by $1$, corresponding to $\lambda_1, \ldots, \lambda_n$, respectively.

We denote $n$ by $n$ eigenvalue matrix and eigenvector matrix by $\mathbf{\Lambda}=\text{diag}(\lambda_1, \ldots, \lambda_n)$ and $\mathbf{S}=\begin{bmatrix}\mathbf{s}_1&\cdots&\mathbf{s}_n\end{bmatrix}$.

Q) Prove that $\mathbf{A}=\mathbf{A}^\text{T}$ if and only if $\mathbf{A}=\mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^\text{T}$, where $\mathbf{Q}\mathbf{Q}^\text{T}=\mathbf{I}$.

A) From $\mathbf{A}=\mathbf{A}^\text{T}$ and $\mathbf{A} = \mathbf{S}\mathbf{\Lambda}\mathbf{S}^{-1}$, \begin{align} \mathbf{A} &= \mathbf{S}\mathbf{\Lambda}\mathbf{S}^{-1}\\ &= (\mathbf{S}\mathbf{\Lambda}\mathbf{S}^{-1})^\text{T}\\ &= (\mathbf{S}^{-1})^\text{T}\mathbf{\Lambda}^\text{T}\mathbf{S}^\text{T}\\ &= (\mathbf{S}^{-1})^\text{T}\mathbf{\Lambda}\mathbf{S}^\text{T}\\ \end{align}

If we assume that $\mathbf{S}=\mathbf{Q}$, then $(\mathbf{S}^{-1})^\text{T}=(\mathbf{Q}^{-1})^\text{T}=(\mathbf{Q}^\text{T})^\text{T}=\mathbf{Q}$.

Hence, $\mathbf{A}=\mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^\text{T}$.

I am very uncomfortable because I used the assumption in the proof. Is this proof enough??

Danny_Kim
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  • It is not enough. However, if you drop the assumption that $A^T=A$, but assume $S=Q$, then you almost have a proof in one direction (use the calculations you have already done, in the right order, to prove that $A^T=Q\Lambda Q^{-1}=A$). Next is to assume $A=A^T$, and show that we must have $S=Q$. – Arthur Dec 02 '17 at 16:44
  • I understood, $\Leftarrow$ is done from my question (with a bit modification), but I have to prove that $\Rightarrow$ again correctly, right? – Danny_Kim Dec 02 '17 at 16:53

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Since $A^T = A$, you can choose the eigenvectors $s_i$ to be an orthogonal set. If you further divide each eigenvector $s_i$ by it's length to obtain $q_i = \frac{s_i}{||s_i||},$ then each $q_i$ will still be an eigenvector, and moreover, matrix $Q$ with columns $q_i$ will satisfy $Q^T = Q^{-1}$. This is because $Q^TQ = I_n$, which holds for any orthogonal matrix $Q$. So for symmetric matrices, diagonalization is always possible and takes the form $A = Q\Lambda Q^T$.

treble
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  • "Since $A^T = A$, you can choose the eigenvectors $s_i$ to be an orthogonal set." This is the statement that I feel someone would need help with, the only one requiring cleverness to discover in one own. The rest of your answer is just mechanic calculation. You could leave it, I'm not saying you should show everything in an answer. It's just a"heads up" to anyone reading this. – Arthur Dec 02 '17 at 16:51
  • Is it okay if we do not prove that your first sentence? "Since $A^\text{T} = A$, we can choose the eigenvectors $\mathbf{s}_i$'s to be orthogonal each other. – Danny_Kim Dec 02 '17 at 16:52
  • @Danny_Kim If I were correcting a test, and your question was on it, someone proving "Since $A^T = A$, you can choose the eigenvectors $s_i$ to be an orthogonal set." And saying "therefore the result follows" might not get more points than this answer, but I would have higher hopes for that student. See my comment above. – Arthur Dec 02 '17 at 16:54
  • @Danny_Kim In a typical linear algebra class, the student would not prove that statement. It's in many linear algebra books without proof. However, the proof that you can do this is part of a classical theorem called the `spectral theorem,' which you can look up if you'd like more information. – treble Dec 02 '17 at 16:55
  • @treble What linear algebra classes have you taken? The proof is really simple to understand (although difficult to discover, as I've pointed out). The relevant lemma is "Eigenvalues corresponding to different eigenvalues of a symmetric matrix are orthogonal". – Arthur Dec 02 '17 at 16:58
  • @Arthur The OP did not assume that the eigenvalues are distinct. If you want to learn more about the spectral theorem, I advise you to ask a question separately. – treble Dec 02 '17 at 17:13
  • @treble I didn't say he did. But equal eigenvalues are handled separately (any eigenvalues corresponding to a single eigenvalues may be given an orthonormal basis, like any other subspace). – Arthur Dec 02 '17 at 17:14
  • @Arthur But not necessarily an orthonormal eigenbasis, unless $A^T = A$, which is the point of the spectral theorem. – treble Dec 02 '17 at 17:17