# Proof for why symmetric matrices are only orthogonally diagonalizable

I am wondering why symmetric matrices are diagonalizable only by orthogonal matrices (and these orthogonal matrices by definition have orthonormal vectors). This is the proof but I don't really get the second part:

Why is this part true:

Finally, by Theorem 7.5, you can conclude that P^{-1}AP is diagonal. So, A is orthogonally diagonalizable.

How do we know it has n linearly independent eigenvectors?

• Your first sentence is false. It’s not that (real) symmetric matrices are only orthogonally diagonalizable, it’s that you can always find such a diagonalization for them. There are many others—just take your orthonormal basis and multiply each vector by an arbitrary nonzero scalar to get another basis that diagonalizes the matrix.
– amd
Oct 1 '18 at 20:06

## 3 Answers

The identity matrix is symmetric, and is diagonalizable by any invertible matrix $$P$$ because $$P^{1}IP=I$$. So such a diagonalization is not necessarily unique.

If $$A$$ is symmetric, then it has an orthonormal basis $$\{ d_1,d_2,\cdots,d_n \}$$ of column eigenvectors with corresponding eigenvalues $$\{ \lambda_1,\lambda_2,\cdots,\lambda_n \}$$. In matrix notation

$$A\left[\begin{array}{cccc}| & | & | & | & | \\ d_1 & d_2 & d_3 & \cdots & d_4 \\ | & | & | & | & | \end{array}\right] \\ = \left[\begin{array}{cccc} | & | & | & | & | \\ \lambda_1 d_1 & \lambda_2 d_2 & \lambda_3 d_3 & \cdots & \lambda_n d_n \\ | & | & | & | & | \end{array}\right] \\ = \left[\begin{array}{ccccc}| & | & | & | & | & \\ d_1 & d_2 & d_3 & \cdots & d_n \\ | & | & | & \vdots & |\end{array}\right] \left[\begin{array}{ccccc} \lambda_1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda_2 & 0 & \cdots & 0 \\ 0 & 0 & \lambda_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda_n \end{array}\right]$$ So $$AU=UD$$ or $$A=UDU^{-1}$$, where $$D$$ is diagonal. The matrix $$U$$ is orthogonal because the columns form an orthonormal basis, thereby forcing $$U^{T}U=I$$.

Conversely, if $$A=UDU^{-1}$$ where $$D$$ is diagonal and $$U$$ is an orthogonal matrix, then every column of $$U$$ is an eigenvector of $$A$$ because $$AU=UD$$.

In the proof of Thm 7.5, the columns vectors $$p_i$$ are the eigenvectors of $$A$$. (You may think about the matrix product on each side of $$AP = PD$$ in a column-wise manner.) Since $$P^{-1}$$ exists at the very beginning of the proof, $$P$$ is nonsingular. This forbids the column vectors $$p_i$$'s from being linearly dependent. (Otherwise, some column $$p_j$$ would be a linear combination of other columns, and this would give $$\det(P) = 0$$, contradicting the invertibility of $$P$$.)

You constructed $$P$$ to have columns being orthonormal eigenvectors. You should know a result stating that an set of orthogonal vectors being linearly independent. So the columns of $$P$$ are your $$n$$ linearly independent eigenvectors.