I'm trying to prove symmetric matrices are orthogonally diagonalizable but I can't go anywhere. In a textbook, a proof is proceed by two steps.
First assume A is n by n symmetric matrix and it has k geometric multiplicity of $\lambda_0$ .
And let $B_0=\{u_1, … , u_k\}$ be basis for eigenspace corresponding to $\lambda_0$ and extend $B=\{u_1, … , u_n\}$ for $R^n$.
Then $AB=B[\lambda I_k \quad X;\; 0 \quad Y ]$. I have understood until here but the textbook said X is zero matrix of $ n\times(n-k)$ which I couldn't go any further.
Second the textbook said considering characteristic
polynomial of $C= [\lambda I_k \quad 0;\; 0 \quad Y ]$ which is similar to A , the algebraic multiplicity of λ0 is the same as the geometric
multiplicity of λ0. But characteristic
polynomial of C is $(\lambda-\lambda_0)^kdet(Y)$ so don't we have to say that the algebraic
multiplicity of λ0 is greater than or equal to the geometric
multiplicity k? I couldn't understand why the algebraic multiplicity and the geometric
multiplicity of λ0 are same.
Thank you for reading this.
Best Answer
If I got your question right, You want to find an orthogonal matrix $X$ such that $X^TAX$ is diagonal. If $\mathbf v_i$ is the i-th eigenvector of $A_{N \times N}$, then $$ A \mathbf v_i = \lambda_i \mathbf v_i $$ And we define $V$ as $V = [\mathbf v_1 , ... , \mathbf v_N]^T$ so we could say: $$ V^TAV = \langle \mathbf v_i,\mathbf v_j\rangle $$ Hence $X=V$ and $V^TAV$ is diagonal matrix.