Fall 2026: Math 590 Quiz 4 Solutions

Problem 1

For \(T: \mathbb{R}^3\to\mathbb{R}^2\) given by \(T(x,y,z) = (3y,\; 2x-5y+4z)\), \(\alpha = \{(1,0,0),\ (1,1,0),\ (1,1,1)\}\) a basis of \(\mathbb{R}^3\) and \(\beta = \{e_1, e_2\}\) a basis of \(\mathbb{R}^2\), find \([T]_\alpha^\beta\). (5 points)

Solution. The columns of \([T]_\alpha^\beta\) are obtained by applying \(T\) to the vectors in \(\alpha\) and expressing the resulting vectors in terms of the basis \(\beta\). Since \(\beta = \{e_1, e_2\}\) is the standard basis of \(\mathbb{R}^2\), the \(\beta\)-coordinates of a vector are just its entries. We compute

\[\begin{align*} T(1,0,0) &= (3\cdot 0,\; 2\cdot 1-5\cdot 0+4\cdot 0) = (0,\,2),\\ T(1,1,0) &= (3\cdot 1,\; 2\cdot 1-5\cdot 1+4\cdot 0) = (3,\,-3),\\ T(1,1,1) &= (3\cdot 1,\; 2\cdot 1-5\cdot 1+4\cdot 1) = (3,\,1). \end{align*}\]

Placing these as the columns gives

\[[T]_\alpha^\beta = \begin{pmatrix} 0 & 3 & 3\\2 & -3 & 1\end{pmatrix}.\]

Problem 2

For the bases \(\alpha = \{e_1, e_2\}\) and \(\beta = \{f_1, f_2\}\), where \(f_1 = \begin{pmatrix} -1\\1\end{pmatrix}\), \(f_2 = \begin{pmatrix} -2\\1\end{pmatrix}\) of \(\mathbb{R}^2\), find the two change of basis matrices and take the product of these matrices. (5 points)

Solution. The columns of \([I]_\beta^\alpha\) are obtained by writing \(f_1, f_2\) in terms of \(e_1, e_2\). Since \(\alpha\) is the standard basis, \(f_1 = -e_1+e_2\) and \(f_2 = -2e_1+e_2\), so

\[[I]_\beta^\alpha = \begin{pmatrix} -1 & -2\\1 & 1\end{pmatrix}.\]

The columns of \([I]_\alpha^\beta\) are obtained by writing \(e_1, e_2\) in terms of \(f_1, f_2\). To do this, we can use Gaussian elimination on the matrices \(\left[\begin{array}{cc|c} f_1 & f_2 & e_1\end{array}\right]\) and \(\left[\begin{array}{cc|c} f_1 & f_2 & e_2\end{array}\right]\). We can also do both eliminations at once by row reducing the augmented matrix \([\,f_1\ f_2\mid e_1\ e_2\,]\) until the left block becomes the identity; the right block then records the coefficients.

\[\begin{aligned} \left[\begin{array}{cc|cc} -1 & -2 & 1 & 0\\1 & 1 & 0 & 1\end{array}\right] &\xrightarrow{-R_1} \left[\begin{array}{cc|cc} 1 & 2 & -1 & 0\\1 & 1 & 0 & 1\end{array}\right] \xrightarrow{R_2-R_1} \left[\begin{array}{cc|cc} 1 & 2 & -1 & 0\\0 & -1 & 1 & 1\end{array}\right]\\[6pt] &\xrightarrow{-R_2} \left[\begin{array}{cc|cc} 1 & 2 & -1 & 0\\0 & 1 & -1 & -1\end{array}\right] \xrightarrow{R_1-2R_2} \left[\begin{array}{cc|cc} 1 & 0 & 1 & 2\\0 & 1 & -1 & -1\end{array}\right]. \end{aligned}\]

The left block is now \(I_2\), so the right block gives \(e_1 = f_1-f_2\) and \(e_2 = 2f_1-f_2\). Hence

\[[I]_\alpha^\beta = \begin{pmatrix} 1 & 2\\-1 & -1\end{pmatrix}.\]

Taking the product we have

\[[I]_\beta^\alpha\,[I]_\alpha^\beta = \begin{pmatrix} -1 & -2\\1 & 1\end{pmatrix}\begin{pmatrix} 1 & 2\\-1 & -1\end{pmatrix} = \begin{pmatrix} -1+2 & -2+2\\1-1 & 2-1\end{pmatrix} = \begin{pmatrix} 1 & 0\\0 & 1\end{pmatrix},\]

and likewise \([I]_\alpha^\beta\,[I]_\beta^\alpha = I_2\). The two change of basis matrices are inverses of each other, so their product is the identity matrix.