# Why does a symmetric matrix have a complete set of eigenvectors and eigenvalues?

I am attempting to learn more about the adjacency matrix(graph theory) but given that I have forgotten a lot of linear algebra, I can't seem to know why this is true. Can someone give me a proof?

• Google is your friend: quandt.com/papers/basicmatrixtheorems.pdf Commented Jun 9, 2015 at 2:50
• Because if $A$ is the matrix in question, an $A$ invariant subspace is also $A^T$ invariant. Commented Jun 9, 2015 at 2:54
• Perhaps this helps: math.stackexchange.com/questions/82467/… Commented Jun 9, 2015 at 5:41
• @Hans Lundmark Ok. So we know "eigenvectors corresponding to a different eigenvalue are orthogonal to each other". But, doesn't the keyword a complete set mean that there are $n$ eigenvalues and one eigenvector corresponding to each of the eigenvalues. I do not think the post says that, but it does say that you can find $n$ orthogonal eigenvectors. Commented Jul 24, 2015 at 2:57
• There are $n$ eigenvalues if you count with them with multiplicity. If they are all distinct, then you have at once a basis with $n$ orthogonal eigenvectors. The difficult thing to show is that if the matrix happens to have a (say) triple eigenvalue, then there really must be a corresponding eigenspace of dimension three (not just one or two), so that you can get three basis eigenvectors from this eigenspace. This is the point of the answer by user level1807 in the question that I linked to. Commented Jul 24, 2015 at 8:11

If a matrix $A$ is diagonalizable, then its minimal polynomial is $$m(\lambda)=(\lambda-\lambda_1)(\lambda-\lambda_2)\cdots(\lambda-\lambda_k)$$ where $\lambda_j$ are the distinct eigenvalues.
Conversely, suppose that the minimal polynomial $m$ for a matrix $A$ factors into distinct linear factors, as written above. Then $$(A-\lambda_l I)\prod_{j\ne l}(A-\lambda_j I)=0.$$ So the non-zero vectors in the range of $\prod_{j\ne l}(A-\lambda_j I)$ are eigenvectors of $A$ with eigenvalue $\lambda_l$. And every vector $x$ can be written as a sum of such vectors because $$1 \equiv \sum_{l=1}^{k}\frac{\prod_{j\ne l}(\lambda-\lambda_j)}{\prod_{j\ne l}(\lambda_l-\lambda_j)}$$ and, hence, $$I = \sum_{l=1}^{k}\frac{1}{\prod_{j\ne l}(\lambda_l-\lambda_j)} \prod_{j\ne l}(A-\lambda_j I)$$ Therefore, a matrix is diagonalizable (equivalently, has a basis of eigenvectors) iff the minimal polynomial for $A$ factors into the product of the distinct linear factors.
A symmetric $A$ has the property that its conjugate transpose $A^{\star}$ is equal to $A$. A symmetric $A$ is a special case of a normal $A$ for which $A^{\star}A=AA^{\star}$. A normal matrix has the property that $$\|Ax\|^{2}=(Ax,Ax)=(A^{\star}Ax,x)=(AA^{\star}x,x)=\|A^{\star}x\|^{2}.$$ Therefore, $A^{2}x=0$ iff $A^{\star}Ax=0$, which implies $Ax=0$ because $$0 = (A^{\star}Ax,x)=(Ax,Ax)=\|Ax\|^{2}.$$ Therefore, the minimal polynomial of a normal $A$ has no repeated factors because $$(A-\lambda_{k}I)^{2}(\cdots)x = 0 \iff (A-\lambda_{k}I)(\cdots)x =0.$$ (Note: $A-\lambda_{k}I$ is normal iff $A$ is normal.)