# Proving $\dim(E_0) \geq n - k$

I found a question from an old exam which I am not really able to wrap my head around. The questions states:

Given $$k and $$v_1, v_2, ..., v_k \in \mathbb{R} ^n$$ non-zero vectors, orthogonal to the standard dot product on $$\mathbb{R}^n$$, where we consider the vectors of $$\mathbb{R}^n$$ as column vectors. Given that $$\lambda_1, \lambda_2, ..., \lambda_k \in \mathbb{R}$$ and $$A = \lambda_1 v_1 v_1^T + \lambda_2 v_2 v_2^T + ... + \lambda_k v_k v_k^T \in \mathbb{R}^{n \times n}$$

Part A asks to prove that $$v_1, v_2, ..., v_k$$ are eigenvectors of $$A$$. I managed to prove this by showing that $$Av_i = \left(\sum_{j=1}^k\lambda_jv_jv_j^T\right)v_i=\sum_{j=1}^k\lambda_jv_j(v_j^Tv_i)=\lambda_iv_i|| v_i||^2=\left(\lambda_i|| v_i||^2\right)v_i$$ $$\forall i = 1, 2, ..., k$$

Part B asks to prove that $$\dim(E_0) \geq n - k$$, where $$E_0$$ the eigenspace is for the eigenvalue $$0$$. I don't really have a direct idea of how to get started with this part.

Part C asks to prove that $$A$$ is diagonalisable and give an orthogonal basis of $$A$$. Again I am not really sure how to get started with this.

Any help is appreciated.

• I think $\Bbb R^{2\times2}$ should be $\Bbb R^{n\times n}$, which presumably denotes the space of $n\times n$ matrices. – Marc van Leeuwen Jan 22 at 17:12

## 2 Answers

For Part B, you can find an orthogonal system $$v_{k+1},v_{k+2},\ldots, v_n\ ,$$ which is orthogonal to $$\{v_1,v_2,\ldots, v_k\}$$ (e.g. by Gram-Schmidt). We can see $$Av_{i}=0$$ for $$k. From this, deduce that $$\dim E_0\ge n-k$$.

Combining the Part A and B, we can see immediately there exists an orthogonal basis consisting of eigenvectors of $$A$$.

Note: Spectral decomposition theorem says every symmetric real matrix is of the same form as that of $$A$$.

Part B: observe that "the eigenspace for the eigenvalue $$0$$" is just a fancy way of saying "null space". Both are the set of all vectors $$v$$ for which $$Av=0$$. I'm assuming you're familiar with computing null spaces?

Part C: a matrix is diagonalizable if the sum of dimensions of its eigenspaces equals the number of columns. For example, a $$3\times 3$$ matrix will be diagonalizable if it has two eigenvalues, and the corresponding dimensions of the eigenspaces are $$1$$ and $$2$$. If we know that a matrix is diagonalizable, diagonalizing it amounts to finding all of the eigenvalue/eigenvector pairs. You can then use Gram-Schmidt to get an orthogonal basis.