Show that for a $n\times n$ symmetric, idempotent matrix $(a_{ij})$, $a_{ii} \ge a_{ii}^2$.

Question

Show that for a $n\times n$ symmetric, idempotent matrix $(a_{ij})$, $a_{ii} \ge a_{ii}^2$. I am not sure whether this result is true.. I am not getting any leads to approach this problem. Anyone?

Answer 1

By spectral decomposition, you can write $$A=\sum_{j=1}^n \lambda_j u_j u_j^T$$ where $\lambda_1,.., \lambda_n$ are eigenvalues and $u_1,.., u_n$ is an Orthonormal basis of eigenvectors.

Next, as $A$ is idempotent, each $\lambda_i$ is 0 or 1.

Finally, if $v= \begin{bmatrix} v_1 \\ ... \\v_n \end{bmatrix}$ is any vector, then $$(vv^T)_{ii}=v_i^2$$

So $a_{ii}$ is simply $$a_{ii}=\sum_{j=1}^n \lambda_j (u_j)_i^2$$ where $(u_j)_i$ denotes the i^th entry in $u_j$.

Now use the fact that the matrix with $u_j$ as columns is othogonal, which gves that the rows of this matrix are an orthonomal basis.

This gives $\sum_{j=1}^n (u_j)_i^2=1$ for each $i,j$.

Therefore $$0 \leq a_{ii}=\sum_{j=1}^n \lambda_j (u_j)_i^2 \leq \sum_{j=1}^n (u_j)_i^2=1$$ which is exactly what you need to prove.