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Let T be the set of all real 2x2 matrices with zero trace: T = {A € Mz(R) tr( A) = 0} a) Prove that T is subspace of V Mz(R). (b) Construct a basis for T. Pro...

Question

Let T be the set of all real 2x2 matrices with zero trace: T = {A € Mz(R) tr( A) = 0} a) Prove that T is subspace of V Mz(R). (b) Construct a basis for T. Prove that your collection does in fact form a basis Hint: You need to decide what typical element of T looks like. What has to be true about the entries along the main diagonal of A for tr(A) = 0?)

Let T be the set of all real 2x2 matrices with zero trace: T = {A € Mz(R) tr( A) = 0} a) Prove that T is subspace of V Mz(R). (b) Construct a basis for T. Prove that your collection does in fact form a basis Hint: You need to decide what typical element of T looks like. What has to be true about the entries along the main diagonal of A for tr(A) = 0?)



Answers

Let $A=\left[\begin{array}{rrrr}1 & 1 & -1 & 1 \\ 2 & -3 & 5 & -6 \\ 5 & 0 & 2 & -3\end{array}\right],$ and let $\mathbf{v}_{1}=$ (-2,7,5,0) and $\mathbf{v}_{2}=(3,-8,0,5)$. (a) Show that $\left\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\}$ is a basis for the null space of $A$. (b) Using the basis in part (a), write an expression for an arbitrary vector $(x, y, z, w)$ in the null space of $A$.

Hello there. Okay. So for this exercise we have Mhm. The basis be defined by these vectors. We want to be three before and we need to find the matrix that represent the transformation well respected. This basis B. Of a linear operator. So this operation is defined as follows. It maps the first vector of the basis on to the second one, the second to the third, the third to the floor and the 4th first. Now the point is that we need to represent these vectors in the basis of the so technically V one Britain in the basis B will be this victory 1000. If you two will be In the same sense will be 0100 B three will be the lecture 0010. and the last one before will be the victor 0001. Of course. All these vectors are the representation in the basis B. Okay, so this is represented in the basis B. B. And big. Now the point is that when we map these vectors using this transformation, what we obtain here in the basis B is the following vector is a vector 0100. In this part. For the two in the basis beat the transformation. The vector rates us 001, zero four B three. This is equals two 0001. And the Before in the basis be will become the vector 100 and zero. Okay, so you can observe that. It's not that hard to grind the equity. The matrix that represent this transformation with respect to the back to the basis be because here we have how it's going to be and transforming, right? So the answer. The final answer. The transformation, the representation of this linear operator in the basis B is the matrix 01 00 00, 001 And 1000.


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