Let $A$ be positive definite and symmetric and let $x$ be the solution of the linear system $$Ax=b$$ Let $0< \lambda_1 \leq \lambda_2 \leq \cdots \leq \lambda_n$ be the eigenvalues of $A$ and $v_1,\dots,v_n$ the corresponding orthonormal eigenvectors. \ 1) Show that $x$ can be written as $$x = \sum_{i=1}^n \lambda_i^{-1}(b\cdot v_i)v_i.$$
My attempt:Since $A$ is positive definite and symmetric, it can be decomposed as $A=Q\Lambda Q^T$, where $Q$ is an orthogonal matrix whose columns are $v_1,\dots,v_n$ and $\Lambda$ is a diagonal matrix whose diagonal entries are $\lambda_1,\dots,\lambda_n$. Then, $$Ax = Q\Lambda Q^Tx = Q\Lambda Q^T\left(\sum_{i=1}^n x_i v_i\right) = Q\left(\sum_{i=1}^n \lambda_i x_i v_i\right) =b$$ where $x_i = (b\cdot v_i)$. Hence, $$x = \sum_{i=1}^n \lambda_i^{-1}(b\cdot v_i)v_i.$$ I'm not sure if I did the $Ax = Q\Lambda Q^Tx$ part correct.
$\begingroup$ $\endgroup$
Add a comment |
1 Answer
$\begingroup$ $\endgroup$
2 You don't need to decompose $A$. Since $Av_i=\lambda_iv_i$ and linearity holds, we have:
$$A \left (\sum_{i=1}^n \lambda_i^{-1} (b \cdot v_i)v_i \right)=\sum_{i=1}^n A(\lambda_i^{-1} (b \cdot v_i)v_i)=\sum_{i=1}^n (b \cdot v_i)\lambda_i^{-1} Av_i=\sum_{i=1}^n(b \cdot v_i)\lambda_i^{-1}\lambda_i v_i \\=\sum_{i=1}^n(b \cdot v_i)v_i=b.$$
Thus, $x=\sum_{i=1}^n \lambda_i^{-1} (b \cdot v_i)v_i$ solves $Ax=b$. If also $Ay=b$, then $A(x-y)=0$. If $x-y \neq0$, then it's an eigenvector of $A$ with eigenvalue $0$, which isn't possible since $A$ is positive definite. Therefore $x-y=0 \Rightarrow x=y$ and our solution is unique.
- $\begingroup$ How would $\sum_{i=1}^{n}(b\cdot v_i)v_i=b?$ $\endgroup$Cole Hendrickson– Cole Hendrickson2023-02-01 15:43:43 +00:00Commented Feb 1, 2023 at 15:43
- 1$\begingroup$ Because $\{ v_i \mid I=1, \ldots , n \}$ is an orthonormal basis. $\endgroup$Robert Shore– Robert Shore2023-02-01 16:15:21 +00:00Commented Feb 1, 2023 at 16:15