Question:medium

If \(T:\mathbb{R}^2\to\mathbb{R}^2\) is a linear transformation defined by \[ T(x,y)=(x,0), \] then the matrix of \(T\) relative to the basis \[ B=\{(0,1),(1,0)\} \] is \(____\).

Show Hint

For a matrix relative to a basis, transform each basis vector and write the resulting coordinate vectors as columns.
  • \(\begin{pmatrix}1& 0\\1& 0\end{pmatrix}\)
  • \(\begin{pmatrix}0& 0\\0& 1\end{pmatrix}\)
  • \(\begin{pmatrix}0& 1\\0& 0\end{pmatrix}\)
  • \(\begin{pmatrix}0& 0\\0& 0\end{pmatrix}\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
We need to find the matrix representation of a linear transformation T with respect to a given basis B. The columns of this matrix are the coordinate vectors of the images of the basis vectors, written with respect to that same basis B.

Step 2: Key Formula or Approach:

Let the basis be \(B = \{\vec{b}_1, \vec{b}_2\}\). The matrix of T relative to B, denoted \([T]_B\), is given by: \[ [T]_B = \begin{pmatrix} [T(\vec{b}_1)]_B & [T(\vec{b}_2)]_B \end{pmatrix} \] where \([T(\vec{b}_i)]_B\) is the coordinate vector of \(T(\vec{b}_i)\) with respect to the basis B. 1. Apply the transformation T to each vector in the basis B. 2. Express each resulting image vector as a linear combination of the basis vectors in B. 3. The coefficients of these linear combinations form the columns of the matrix \([T]_B\).

Step 3: Detailed Explanation:

The basis is \(B = \{\vec{b}_1, \vec{b}_2\}\) where \(\vec{b}_1 = (0,1)\) and \(\vec{b}_2 = (1,0)\). The transformation is \(T(x,y)=(x,0)\). 1. Transform the basis vectors:
- For the first basis vector \(\vec{b}_1 = (0,1)\): \[ T(\vec{b}_1) = T(0,1) = (0,0) \] - For the second basis vector \(\vec{b}_2 = (1,0)\): \[ T(\vec{b}_2) = T(1,0) = (1,0) \] 2. Find the coordinates of the images with respect to B:
- For \(T(\vec{b}_1) = (0,0)\): We need to find scalars \(c_1, c_2\) such that \((0,0) = c_1\vec{b}_1 + c_2\vec{b}_2 = c_1(0,1) + c_2(1,0)\). This gives \((0,0) = (c_2, c_1)\), so \(c_1=0\) and \(c_2=0\). The coordinate vector is \([T(\vec{b}_1)]_B = \begin{pmatrix} 0 0 \end{pmatrix}\). This will be the first column of our matrix. - For \(T(\vec{b}_2) = (1,0)\): We need to find scalars \(d_1, d_2\) such that \((1,0) = d_1\vec{b}_1 + d_2\vec{b}_2 = d_1(0,1) + d_2(1,0)\). The vector \((1,0)\) is exactly our second basis vector \(\vec{b}_2\). So, \((1,0) = 0 \cdot (0,1) + 1 \cdot (1,0) = 0\vec{b}_1 + 1\vec{b}_2\). This gives \(d_1=0\) and \(d_2=1\). The coordinate vector is \([T(\vec{b}_2)]_B = \begin{pmatrix} 0 1 \end{pmatrix}\). This will be the second column of our matrix. Note on Discrepancy: My second column is \((0,1)\). Let me recheck. \(T(1,0) = (1,0)\). We need to write this in the basis B. \((1,0) = d_1(0,1) + d_2(1,0)\). The vector \((1,0)\) is \(\vec{b}_2\). Wait, the basis is given as B=\(\{(0,1), (1,0)\}\). Let \(\vec{e}_1 = (0,1)\) and \(\vec{e}_2 = (1,0)\). \(T(\vec{e}_1)=T(0,1)=(0,0) = 0\vec{e}_1 + 0\vec{e}_2\). First column is \(\begin{pmatrix}00\end{pmatrix}\). \(T(\vec{e}_2)=T(1,0)=(1,0) = \vec{e}_2 = 0\vec{e}_1 + 1\vec{e}_2\). Second column is \(\begin{pmatrix}01\end{pmatrix}\). Matrix is \(\begin{pmatrix} 0 & 0 0 & 1 \end{pmatrix}\). This matches option (B). Let me re-read the provided solution which is (C). Maybe the order of basis vectors in the matrix is different? The convention is \(j\)-th column is \(T(\vec{b}_j)\). The marked answer is \(\begin{pmatrix} 0 & 1 0 & 0 \end{pmatrix}\). This would mean the first column is \(\begin{pmatrix}00\end{pmatrix}\) and the second is \(\begin{pmatrix}10\end{pmatrix}\). First column: \(T(\vec{b}_1)=(0,0)\), which is \(0\vec{b}_1+0\vec{b}_2\). Correct. Second column: \(T(\vec{b}_2)\) has coordinates \(\begin{pmatrix}10\end{pmatrix}\). This means \(T(\vec{b}_2) = 1\vec{b}_1 + 0\vec{b}_2 = \vec{b}_1 = (0,1)\). But we calculated \(T(\vec{b}_2) = T(1,0) = (1,0)\). So \(T(\vec{b}_2)=(1,0)\), but the coordinate vector required for matrix (C) is \(\begin{pmatrix}10\end{pmatrix}\), meaning \(T(\vec{b}_2)\) should be \(\vec{b}_1=(0,1)\). This is a contradiction. Let's try the standard basis \(\{(1,0),(0,1)\}\). \(T(1,0)=(1,0)\), \(T(0,1)=(0,0)\). The matrix would be \(\begin{pmatrix} 1 & 0 0 & 0 \end{pmatrix}\). There seems to be an error in the question or the provided answer key. Let's check the basis order one more time. Basis is \(B = \{\vec{b}_1=(0,1), \vec{b}_2=(1,0)\}\). Image of first basis vector: \(T(\vec{b}_1) = T(0,1) = (0,0)\). Coordinates of image: \((0,0) = 0\vec{b}_1 + 0\vec{b}_2\). So first column is \(\begin{pmatrix} 0 0 \end{pmatrix}\). Image of second basis vector: \(T(\vec{b}_2) = T(1,0) = (1,0)\). Coordinates of image: \((1,0) = c_1\vec{b}_1 + c_2\vec{b}_2 = c_1(0,1) + c_2(1,0) = (c_2, c_1)\). So \(c_2=1, c_1=0\). Second column is \(\begin{pmatrix} 0 1 \end{pmatrix}\). Resulting matrix: \(\begin{pmatrix} 0 & 0 0 & 1 \end{pmatrix}\). This is option (B). The key is wrong.

Step 4: Final Answer:

The correct matrix representation is \(\begin{pmatrix} 0 & 0 0 & 1 \end{pmatrix}\), which is option (B). The provided answer key for option (C) is incorrect.
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