# Does $A$ is similar to a symmetric matrix $\implies$ $A$ is symmetric?

Let $A\in \mathbb R^{n\times n}$. Is it true that

$A$ similar to a symmetric matrix $\implies$ $A$ symmetric ?

Let $B$ symmetric s.t. $A=PBP^{-1}$. Then $$A^T=(P^{-1})^TB^T P^T=(P^{-1})^T BP^T.$$

For me there is no reason that $P$ is orthogonal, so I would say it's false a priori. But in the same time, this theorem should be true since operator is self adjoint $\iff$ it's diagonalizable. I also know that matrices in any basis of Self Adjoint operator are symmetric. But if A is similar to a symmetric matrix, then it's diagonalizable and thus self adjoint, and thus, it should be symmetric in any basis... this is wrong ? If yes, why ?

• You are right, it is not true. – gj255 Jul 26 '18 at 11:30
• Better to find counterexample; since if this is the case then only symmetric matrix are diagonalizable. – maths student Jul 26 '18 at 11:31
• It is false that an operator is self-adjoint if and only if it is diagonalizable. – egreg Jul 26 '18 at 12:15

Try a $P$ that is not orthogonal.

Let me try $P=\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}$, $P^{-1}=\begin{bmatrix} \frac12 & 0 \\ 0 & 1\end{bmatrix}$

Let $B=\begin{bmatrix} 1 & 1 \\ 1 & 1\end{bmatrix}$

then $$A=PBP^{-1}=\begin{bmatrix} 2 & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 1\end{bmatrix}\begin{bmatrix} \frac12 & 0 \\ 0 & 1\end{bmatrix}=\begin{bmatrix} 2 & 0 \\ 0 & 1\end{bmatrix}\begin{bmatrix} \frac12 & 1\\ \frac12 & 1\end{bmatrix}=\begin{bmatrix} 1 & 2\\ \frac12 & 1 \end{bmatrix}$$

which is not symmetrical.

• Thank you. I have a doubt with something : I know that an operator is self adjoint $\iff$ it's diagonalizable. I also know that matrices in any basis of Self Adjoint operator are symmetric. But if $A$ is similar to a symmetric matrix, then it's diagonalizable and thus self adjoint, and thus, it should be symmetric in any basis... this is wrong ? – MathBeginner Jul 26 '18 at 11:42
• @MathBeginner A matrix of a self-adjoint operator is symmetric in every orthonormal basis, but not necessarily in every basis. – Adayah Jul 26 '18 at 11:43
• @Adayah: And : if there is an orthonormal basis such that $A$ is symmetric, then it's symmetric in every orthonormal basis ? – MathBeginner Jul 26 '18 at 11:45
• @MathBeginner Yes. – Adayah Jul 26 '18 at 11:46
• @MathBeginner The reason is, that a change of basis from an orthonormal basis to another orthonormal basis is an orthogonal matrix (Easiest way to see that: change of basis from standard basis to orthonormal basis has the vectors of the basis in its columns $\Rightarrow$ is orthogonal). So if you change the basis, the resulting matrix $P^\top AP$ is still symmetric (since $(P^\top A P )^\top = P^\top A^\top P^{\top\top}$) – Babelfish Jul 26 '18 at 12:02

Take any $A$ which is diagonizable but not symmetric. So $A = TDT^{-1}$ for a diagonal matrix $D$ and invertible $T$. For any orthogonal matrix $O$ $A$ is similar to $ODO^\intercal$, which is symmetric, but $A$ is not.

Since there is already a counterexample in the other answer, I will focus on this:

But in the same time, this theorem should be true since symmetric matrices looks very important, and such a result with this type of matrix should hold...

It is true that symmetric matrices (with real entries) are important because they correspond to self-adjoint linear operators. Such operators are always* diagonalizable in an orthonormal basis and have real eigenvalues. Moreover, those attributes characterize self-adjoint operators, i.e. any operator diagonalizable in an orthonormal basis with real eigenvalues is self-adjoint.

So basically a symmetric matrix is important because of its properties which relate to the inner product. But similarity of matrices does not preserve the inner product structure (only conjugation with unitary matrices does), hence it also does not necessarily preserve symmetric matrices. A similar example in topology is that a space homeomorphic to a complete metric space need not be complete.

*In this answer I am only referring to finite-dimensional unitary spaces.

An operator over $\mathbb{R}^n$ is self-adjoint if and only if it is diagonalizable with an orthogonal matrix.

You are missing the clause shown in italics.

For instance, the matrix \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} is diagonalizable, having two distinct eigenvalues ($1$ and $2$). Hence the matrix is similar to a diagonal, hence symmetric, matrix. However, this matrix is not orthogonally similar to a diagonal matrix, because its basic eigenvectors are not orthogonal to each other.

If $A$ is diagonalizable with an orthogonal matrix, that is $$A=PDP^T$$ with $D$ diagonal and $P^T=P^{-1}$, then $$A^T=(PDP^T)^T=PD^TP^T=PDP^T=A$$ The converse ($A$ symmetric is diagonalizable with an orthogonal matrix) is a deeper theorem, a consequence of the more general spectral theorem.