# Diagonalizing a matrix that is not upper-triangular

I was reading through Linear Algebra Done Right and came across example 5.40 in chapter 5.C: Define $$T$$ as an operator on $$\mathbb{R}^2$$ as $$T(x, y) = (41x + 7y, -20x + 74y)$$. Clearly the matrix of $$T$$ w.r.t the standard basis is $$\begin{pmatrix} 41 & 7\\ -20 & 74\\ \end{pmatrix}$$ which is not diagonal. Axler provides that the matrix w.r.t the basis $$\{(1, 4), (7, 5)\}$$ is $$\begin{pmatrix} 69 & 0\\ 0 & 46\\ \end{pmatrix}$$ which makes $$T$$ diagonalizable. I was just wondering how he found a basis where $$T$$ is diagonalizable without using determinants. Would anybody care to explain this process?

• This is a very important and standard topic in linear algebra that perhaps comes later in the text? Those vectors are called eigenvectors and the values on the diagonal are called eigenvalues. The process of finding the eigenbasis and conjugating the matrix for $T$ to the diagonal one is called diagonalization. Perhaps, at this moment, you are meant to just check that $T$ does in fact have that diagonal matrix in that particular basis. Mar 28, 2022 at 18:01
• The characteristic equation of the matrix is written and solved to find the eigenvalues as 69 and 46. Then the eigen vectors are found to get the required basis. Mar 28, 2022 at 22:22

## 1 Answer

I would say he began with the diagonal matrix 3,2 (as being different from the identity) and intended basis as suggested above, with necessary $$\frac{1}{23} \left( \begin{array}{rr} 1&7 \\ 4&5 \\ \end{array} \right) \left( \begin{array}{rr} -5&7 \\ 4&-1 \\ \end{array} \right) = \left( \begin{array}{rr} 1&0 \\ 0&1 \\ \end{array} \right)$$

Then he would have some fractions remaining in $$\frac{1}{23} \left( \begin{array}{rr} 1&7 \\ 4&5 \\ \end{array} \right) \left( \begin{array}{rr} 3&0 \\ 0&2 \\ \end{array} \right) \left( \begin{array}{rr} -5&7 \\ 4&-1 \\ \end{array} \right) = \frac{1}{23} \left( \begin{array}{rr} 41&7 \\ -20&74 \\ \end{array} \right)$$ So he could multiply the diagonal matrix by $$23$$ to find $$\frac{1}{23} \left( \begin{array}{rr} 1&7 \\ 4&5 \\ \end{array} \right) \left( \begin{array}{rr} 69&0 \\ 0&46 \\ \end{array} \right) \left( \begin{array}{rr} -5&7 \\ 4&-1 \\ \end{array} \right) = \left( \begin{array}{rr} 41&7 \\ -20&74 \\ \end{array} \right)$$