[Math] Prove that the largest singular value of a matrix is greater than the largest eigenvalue

linear algebra

Let $\sigma_1$ be the largest singular value of the matrix $A = (a_{ij})$. Show that $\sigma_1 >= \lambda_{max}$, where $\lambda_{max}$ denotes the largest eigenvalue of $A$, and that $\sigma_1 \geq |a_{ij}|_{max}$.

$A$ must be a square matrix, otherwise it would not have any eigenvalues. One of the problems that I have worked so far includes a square matrix, and the above statement holds true for that problem.

I am unsure of how to approach proving this for the general case however. I know that $\sigma_1 = \sqrt{\lambda_{max}(A^TA)}$.

Is there a relationship between $\lambda_{max}(A^TA)$ and $\lambda_{max}(A)$ that can be leveraged to prove this?

Best Answer

The idea mainly comes from 'Matrix analysis' in chapter 5 by Roger A.Horn

Let A be the matrix we will study

  1. prove the spectral radius ≤ the matrix norm.
  2. the spectral norm defined as the largest singular value of A is also a type of matrix norm.
  3. from 1 and 2, the spectral norm ≥ spectral radius

so $$\sigma_{max} ≥ |\lambda_{max}|$$

and if A is nonsingular, $$\sigma_{min} ≤ |\lambda_{min}|$$

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