# Finding a matrix corresponding to a linear transformation for a variable domain

Assume that the first two columns $$A_1$$ and $$A_2$$ of the following matrix $$A = \begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix}$$ are linearly independent so that $$\bar{\mathcal{B}}=\{A_1,A_2\}$$ forms an ordered basis for $$\mathbb{R}^2$$. Let $$\{X_0\}$$ be a basis of the nullspace of $$A$$. Then $$\mathcal{B}=\{[1, 0, 0]^t, [0, 1, 0]^t, X_o\}$$ is a linearly independent set. Prove that the matrix of $$T_A:\mathbb{R}^3 \rightarrow \mathbb{R}^2$$ with respect to the ordered bases $$\mathcal{B}$$ and $$\bar{\mathcal{B}}$$ is $$M = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}.$$

I'm able to understand why the third column of $$M$$ is a zero column vector, as $$X_0$$ belongs in the nullspace of $$A$$, so $$T_A(X_0) = [0, 0]^t$$, but I'm not sure how the other two columns are derived.

• Hint: For any basis $\{v_1,v_2\}$ whatsoever, what are the coordinates of $v_1$ relative to this basis?
– amd
Jul 23, 2019 at 0:42

To find the matrix that represents $$\textsf{T}_A$$ with respect to the basis $$\mathcal B =\{e_1,e_2,x_0\}$$ and $$\bar{ \mathcal{B}}=\{A_1,A_2\}$$ first you get the image (under the transformation) of each of the elements in $$\mathcal B$$ and then you put them as a linear combination of the elements that is living in the second basis, $$\bar{ \mathcal{B}}$$.
In this case $$\textsf{T}_A (e_1)=Ae_1 = \begin{pmatrix} a\\d \end{pmatrix}=1A_1+0A_2$$ $$\textsf{T}_A (e_2)=Ae_2 = \begin{pmatrix} b\\e \end{pmatrix}=0A_1+1A_2$$ $$\textsf{T}_A (x_0)=Ax_0 = \begin{pmatrix} 0\\0 \end{pmatrix}=0A_1+0A_2$$ And finally you put the coefficients as the columns of your matrix. That is $$M_1=(1,0)^t$$ $$M_2=(0,1)^t$$ $$M_3=(0,0)^t$$