Consider T ∈ℒ(𝔽^3) defined by T(x, y, z) = (2x + y, 2y, y +...

Consider $T \in \mathcal{L}(\mathbb{F}^3)$ defined by $T(x, y, z) = (2x + y, 2y, y + 2z)$, $\forall (x, y, z) \in \mathbb{F}^3$ (a) Find the matrix of $T$ with respect to the standard basis of $\mathbb{F}^3$. (b) Assuming that $\lambda = 2$ is the only eigenvalue of $T$, explain why $T$ is not diagonalisable. (c) Find a basis with respect to which the matrix of $T$ is upper triangular.

Transcript

00:02 We're given a transformation t from r3 to itself defined by t of x, y, z equals 2x plus y plus z, right, minus 2 z, 2x plus 3y minus 4 z.

00:42 And it's too much.

00:43 It's like this would be so funny to me.

00:46 It just seems like a fake joke.

00:48 And then x plus y minus z.

00:52 I don't even know any d.

00:56 We're asked to find all the eigenvalues of t and to find a basis of each eigenspace.

01:07 And to find a basis of each eigenstace we're asked if the transformation t is, is diagonalizable, and if so, to find a basis s of r3 that diagonalizes t and to find its diagonal representation d.

01:27 To answer this question, first let's find a matrix a that represents t relative to the usual basis of r3.

01:35 We'll do this by writing down the coefficients of x, y, and z as rows, and then we'll find the characteristic polynomial of this matrix a, which happens to also be the characteristic polynomial of t.

01:51 That a, which is a matrix of t relative to the standard basis.

02:00 Well, this is the three by three matrix two, one, negative two, two, three, negative four, and one, one, negative two.

02:16 Oh, sorry, it's just negative one, that's a z minus six.

02:28 And this is a 3x3x3 matrix, so to calculate the characteristic polynomer.

02:32 I'm going to need a few things.

02:34 The trace of this matrix a is 2 plus 3 minus 1, which is 4.

02:41 The determinants of this matrix a, i'm not going to do the calculation here.

02:45 It's fairly standard, and you'll get 2.

02:50 The co -factor a1 -1, this is the determinant of 3 -4 -1 -1 -negative 1, which is 1.

03:01 Co -factor a -2 -2 -2 -1 -1, which is 0.

03:08 And the co -factor a -3 is the determinant of the sub -matrix 2 -1 -2 -1, which is 4.

03:17 And therefore, it follows to be some of these co -factors, a -i -i, is 5.

03:26 And now we use the formula for the characteristic polynomial of a 3x3 matrix.

03:32 So we have delta of t is t cubed minus the trace of a, which was 4 times t squared, plus the sum of the cofactors, 5 times t minus the determinant of a, which is 2.

03:54 This is a cubic function which we can factor...

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