Let A=[ 1 1 -1 3 ] and ℬ={𝐛1, 𝐛2}, for 𝐛1=[ 1 1 ] 𝐛2=[ 5 4 ] . Defin

step 1 We're given a matrix A and a basis B, where B is the vectors B1 and V2. And we're given a linear

Let A=[ 1 1 -1 3 ] and ℬ={𝐛1, 𝐛2}, for 𝐛1=[ ...

Let $A=\left[\begin{array}{rr}{1} & {1} \\ {-1} & {3}\end{array}\right]$ and $\mathcal{B}=\left\{\mathbf{b}_{1}, \mathbf{b}_{2}\right\},$ for $\mathbf{b}_{1}=\left[\begin{array}{l}{1} \\ {1}\end{array}\right]$ $\mathbf{b}_{2}=\left[\begin{array}{l}{5} \\ {4}\end{array}\right] .$ Define $T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ by $T(\mathbf{x})=A \mathbf{x}$ a. Verify that $\mathbf{b}_{1}$ is an eigenvector of $A$ but $A$ is not diagonalizable. b. Find the $\mathcal{B}$ -matrix for $T$

Let $A=\left[\begin{array}{rr}{1} & {1} \\ {-1} & {3}\end{array}\right]$ and $\mathcal{B}=\left\{\mathbf{b}_{1}, \mathbf{b}_{2}\right\},$ for $\mathbf{b}_{1}=\left[\begin{array}{l}{1} \\ {1}\end{array}\right]$ $\mathbf{b}_{2}=\left[\begin{array}{l}{5} \\ {4}\end{array}\right] .$ Define $T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ by $T(\mathbf{x})=A \mathbf{x}$
a. Verify that $\mathbf{b}_{1}$ is an eigenvector of $A$ but $A$ is not diagonalizable.
b. Find the $\mathcal{B}$ -matrix for $T$

Key Concepts

Eigenvectors and Eigenvalues

An eigenvector of a linear transformation is a nonzero vector that, when the transformation is applied, results in a scalar multiple of itself; the scalar is called the eigenvalue. In the context of a matrix, verifying that a vector is an eigenvector involves checking whether multiplying the matrix by the vector yields a result that is simply the eigenvalue times the original vector.

Diagonalizability

A matrix is diagonalizable if it has enough linearly independent eigenvectors to form a basis of the vector space. This means that the matrix can be expressed in a form where all off-diagonal elements are zero under a suitable change of basis. When a matrix does not have a complete set of linearly independent eigenvectors, as in the case when an eigenvalue has geometric multiplicity lower than its algebraic multiplicity, the matrix is not diagonalizable.

Matrix Representation with Respect to a Basis

Changing the matrix representation of a linear transformation to a different basis involves expressing the transformation in terms of the new coordinate system. This process typically requires computing the coordinates of the transformed basis vectors relative to the new basis. The resulting matrix, often referred to as the 'B-matrix', encapsulates the action of the transformation in the new basis, and is obtained through a similarity transformation if the original matrix is expressed in the standard basis.

Transcript

00:01 We're given a matrix a and a basis b, where b is the vectors b1 and v2.

00:14 And we're given a linear transformation t from r2 to r2, which is defined by t of x equals ax.

00:24 In part a, we're asked to verify that the vector b1 is an eigenvector of a, but that a is not diagonalizable.

00:41 So this is pretty easy to verify it's an eigenvector of a.

00:45 We'll simply calculate a times b1.

00:50 Now we're given that a is the matrix, the column vector is 1 -1 -13, and that b -1 is the vector 1 -1.

01:05 So performing matrix multiplication, we get 1 plus 1 is 2, and negative 1 plus 3 is also 2, which is equal to 2 times 1 -1, which is in turn equal to 2 times b1.

01:23 So clearly, b1 is an eigenvector of a with an eigenvalue of 2.

01:42 Now, to show that a is not diagonalizable, we need to find the eigenvalues of a.

01:49 So we have that the characteristic polynomial of a.

01:56 This is going to be the determinant of a minus i, a minus lambda i, i mean.

02:02 So this is going to be 1 minus lambda times 3 minus lambda.

02:14 Minus negative 1 or plus 1.

02:18 So we get lambda squared minus 4 lambda plus 3 plus 1 is plus 4.

02:30 And this can be factored as lambda minus 2 squared.

02:35 So we see that a has the eigenvalue lambda equals 2 with multiplicity 2.

02:48 That's its only eigenvalue.

02:59 Now we have that the matrix a minus 2i, well this is the matrix, negative 1, 1, negative 1, 1...