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In Exercises $13-16,$ define $T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ by $T(\mathbf{x})=A \mathbf{x}$ . Find a basis $\mathcal{B}$ for $\mathbb{R}^{2}$ with the property that $[T]_{\mathcal{B}}$ is diagonal.$$A=\left[\begin{array}{rr}{0} & {1} \\ {-3} & {4}\end{array}\right]$$

$b_{1}=\left[\begin{array}{l}{1} \\ {1}\end{array}\right], b_{2}=\left[\begin{array}{l}{1} \\ {3}\end{array}\right]$

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 4

Eigenvectors and Linear Transformations

Vectors

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So we're given a transformation t in terms of a matrix A and were asked to find a basis be for our two with the property at the matrix of tea with respect to the basis be is diagonal and we're given that the Matrix A in this case is the matrix with column vectors zero negative 314 This means if this is a in the characteristic polynomial of a is going to be the determinant of a minus slammed I So this is going to be negative Lambda Times for minus lambda and then minus negative three or plus three. This is the same as Lambda Squared minus four Lambda plus three which could be factored as lambda minus one times lambda minus three. So we see that since this is the characteristic polynomial, the Eigen values of a are lambda one equals one and Lambda two equals three. This is when the characteristic polynomial zero. So if lambda is equal to one well, we calculate that the matrix a minus slander I which is simply a minus. I This is going to be the matrix. Negative 11 negative three three. And so the equation a minus I times Vector X equals zero. This is equivalent to the system with the single equation Negative x one plus X two equals zero. So we have that X one equals x two, where X two is a degree of freedom. So a basis specter for this Eigen space, for example, we could take to be V one equals if we take X to be one 11 Now consider the egg in space or lambda is equal to three. In this case, we have re matrix a minus three I is given by negative 31 negative three one And we see that the mate the equation a minus three i times X equals zero is equivalent to solving the single equation. Negative three x one plus X two equals zero Or, in other words, we have that X one is equal to one third x two where X to of course is free. We have one degree of freedom. So, for example, if we take X two to be three, then we have that The vector V two, which is one third of three which is 13 is a basis Fekter For this Eigen space, it's nice because it has thes whole numbers which are easy to work with. Now, from these two Eigen vectors, we can construct the Matrix P, which has V one V two as column vectors. And so this is the Matrix with column vectors 11 and 13 And we know that this diagonal eyes is a so by a previous the're, um we have that the basis v one be to has the property the bee matrix of the transformation from X two x is a diagonal matrix.

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