I am unable to prove the following fact regarding matrices:

  1. If $A$ is a symmetric matrix then there exists a lower triangular matrix $T$ with non-negative diagonal entries such that $A=TT^t$ where $T^t$ denotes transpose of $T$.
  2. The largest eigen value of $T^tT$ exceeds or equals the largest diagonal entry of $T^tT$.

For $1$ it will be sufficient if I could know whether there exists some standard theorem on it as I am unable to prove it.

For $2$.I started with a contradiction that the largest eigen value of $T^tT$ is less than the largest diagonal entry of $T^tT$. Then all the eigen values of $T^tT$ is less than the largest diagonal entry of $T^tT$.

But how to arrive at a contradiction from here?

Please help.


1 Answer 1


For 1, as pointed out by @Emilio, cholesky decomposition.

For 2, Let eigenvalue decomposition of $T^TT=W\Sigma W^T$

$$\max_{\left\|v\right\|=1}v^TT^TTv=\max_{\left\|v\right\|=1}v^TW\Sigma W^Tv=\max_{\left\|y\right\|=1}y\Sigma y=\max_{\left\|y\right\|=1}\sum_{i=1}^n\lambda_iy_i^2\leq \lambda_{max}$$

Notice that the $i$-th diagonal entry of $T^TT=e_i^TT^TTe_i$, where $e_i$ is the standard $i$-th unit basis which satisfies the constraint $\left\|e_i\right\|=1$.

Hence the largest eigenvalue is at least the largest diagonal entries.


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