Alireza Kazemipour

Alireza Kazemipour

PhD student at the University of Alberta

Some reminders on Linear Algebra

1 minute read

Published:

In this post I’ll mention a handful reminders of Linear Algebra.

  • We know that for an eigenvalue $\lambda$ of a matrix $A$ we have that $Ax = \lambda x$, or equivalently $(A - \lambda I)x = 0$. To find $\lambda$, we solve $\mathrm{det}(A - \lambda I) = 0$. But, why?

The reason is that in order to find non-zero solutions of $(A - \lambda I)x = 0$, $(A - \lambda I)$ must be non-invertible, hence its determinant should be zero!

Facts:

  • $\mathrm{trace}(A) = \lambda_1 + \lambda_2 + \dots$

  • $\mathrm{det}(A) = \lambda_1 \cdot \lambda_2 \dots$

  • Why $\mathrm{Det}(A) = \mathrm{Det}(A^\top)$ for any matrix $A$?

We know that $PA = LU$ (using factorization by elimination), where $P$ is a permutation matrix (each row only has one element equal to one and the rest are zero), $L$ is a lower triangular matrix with its diagonal all equal to one, and $U$ is an upper triangular matrix with the pivots of $A$ on its diagonal. For$(PA)^\top = (LU)^\top$, we have that $A^\top P^\top = U^\top L^\top$. So,

\[\begin{align*} \mathrm{Det}(PA) & = \mathrm{Det}(P)\cdot \mathrm{Det}(A) = \mathrm{Det}(L)\cdot \mathrm{Det}(U), \quad \mathrm{and} \\ \mathrm{Det}\left((PA)^\top\right) & = \mathrm{Det}\left(A^\top\right) \cdot \mathrm{Det}\left(P^\top\right) = \mathrm{Det}\left(U^\top\right)\cdot \mathrm{Det}\left(L^T\right). \end{align*}\]

We know since $L, U$ are triangular $\mathrm{Det}(L), \mathrm{Det}(U) = \mathrm{Det}\left(L^\top\right), \mathrm{Det}\left(U^\top\right)$. If $\mathrm{Det}(P) = \mathrm{Det}\left(P^\top\right)$, then it must be the case that $\mathrm{Det}(A) = \mathrm{Det}\left(A^\top\right)$. Now we prove that $\mathrm{Det}(P) = \mathrm{Det}\left(P^\top\right)$:

Let $\pi \in S_n$ be a permutation and $P_\pi$ be the permutation matrix that is obtained by applying $\pi$ to the identity matrix. Then, $(P_\pi)_{ij} = 1$ if $i = \pi(j)$, and otherwise zero. Using Leibniz formula we have:

\[\mathrm{Det}(P_\pi) = \sum_{\sigma \in S_n}\mathrm{sgn}(\sigma) \Pi_{i = 1}^n (P_\pi)_{i\sigma(i)}.\]

They only way the inner product is non-zero is that if $\sigma = \pi$. Hence, $\mathrm{Det}(P_\pi) = \mathrm{sgn}(\pi)$. $\mathrm{sgn}(\pi)$ is equal to $(-1)^k$, where $k = \sum_{i = 1}^n \mathbb{I}${$i \neq \pi(i)$}. Hence, $\mathrm{Det}(P_\pi) = \pm 1$ . Using the identity $PP^\top = I$, we have that $\mathrm{Det}(P)\cdot\mathrm{Det}(P^\top) = 1$. Hence, $\mathrm{Det}(P)$ must be equal to $\mathrm{Det}\left(P^\top\right)$, so the identity holds. $\square$

Reference