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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.

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

1 Answer 1

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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 $$M_1=(1,0)^t$$ $$M_2=(0,1)^t$$ $$M_3=(0,0)^t$$

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