# Question about proving symmetric matrices are diagonalizable

Definition : If a n by n matrix $A$ is orthogonally congruent to another matrix $B$, then there exist an orthogonal matrix $C$ such that $$A = C^{-1}BC$$

Theorem: If $A$ is symmetric, then $A$ is orthogonally congruent to a diagonal matrix, $B = diag (\lambda_1 , .....,\lambda_n)$ where $\lambda_1,....\lambda_n$ are the (necessarily real) eigenvalues of $A$ (with multiple eigenvalues repeated in accordance with their multiplicities)

Why is this true?? If $A$ has n distinct eigenvalues its obvious that this is true, for we can have, $$C = col (\bf{v_1},v_2,....v_n)$$ where $\bf{v_1,v_2...,v_n}$ are all the eigenvectors associated with the eigenvalues, and $$AC = CB \\ B = diag(\lambda_1,....\lambda_n) \\ A = C^{-1}BC$$ But if we do the same thing for the case where the eigenvalues are repeated, then wouldnt $C$ have multiple columns of the linearly dependent vectors? (If we use the repeated eigenvalues to solve for the eigenvectors wouldnt they be the same?) If so, $det(C) = 0$ then how isit still possible that $$A = C^{-1}BC$$ when $C^{-1}$ doesnt exist??

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Basically, If $A$ is symmetric and $v$ is an eigenvector of $A$, then we have that the orthogonal complement to $\left<v\right>$ is invariant under multiplication by $A$. Check out page 17 here: maths.ox.ac.uk/system/files/coursematerial/2013/2632/18/… –  JC574 Jul 23 '14 at 15:09
Dear CodeBreaker: I retitled the question so that people would not automatically ignore it, and so that there would be a remote chance someone with a similar question would be able to find it. Please consider taking the same things into consideration when titling your next question! Regards –  rschwieb Jul 23 '14 at 15:17

• Just because we have $n$ linearly independent vectors, doesn't mean that they form an orthonormal basis. If $A$ were any matrix with $n$ distinct eigenvalues, then $C = col(v_1, \dots, v_n)$ would have to be invertible, but is not necessarily orthogonal.
• Some matrices have eigenspaces that are more than $1$-dimensional. As an example, $\pmatrix{1&0&0\\0&1&0\\0&0&1}$ has only one eigenvalue, but certainly has a basis of eigenvectors.