# The Rank and Nullity of a Linear Transformation from Vector Spaces of Matrices to Polynomials ## Problem 676

Let $V$ be the vector space of $2 \times 2$ matrices with real entries, and $\mathrm{P}_3$ the vector space of real polynomials of degree 3 or less. Define the linear transformation $T : V \rightarrow \mathrm{P}_3$ by
$T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = 2a + (b-d)x – (a+c)x^2 + (a+b-c-d)x^3.$

Find the rank and nullity of $T$. Add to solve later

## Solution.

To find the rank of $T$, we will use its matrix representation relative to the standard bases for $V$ and $\mathrm{P}_3$. Define the matrices
$\mathbf{e}_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} , \, \mathbf{e}_2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} , \, \mathbf{e}_3 = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} , \, \mathbf{e}_4 = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$ and the polynomials
$p_1(x) = 1 , \quad p_2(x) = x , \quad p_3(x) = x^2 , \quad p_4(x) = x^3.$

Then the standard bases are
$B = \{ \mathbf{e}_1 , \mathbf{e}_2 , \mathbf{e}_3 , \mathbf{e}_4 \} \subset V \mbox{ and } C = \{ p_1 , p_2 , p_3 , p_4 \} \subset \mathrm{P}_3.$

We now find the matrix of representation of $T$ relative to these bases. This will be a $4 \times 4$ matrix whose $i$-th row is the column vector of $T(e_i)$ relative to the basis $C$. For example, we have
$T(e_1) = 2 – x^2 + x^3 = 2p_1 – p_3 + p_4 ,$ and so the first column of the matrix is the column vector
$[T(e_1)]_{C} = \begin{bmatrix} 2 \\ 0 \\ -1 \\ 1 \end{bmatrix} .$

This vector becomes the first column of the matrix. After performing this process for the other three basis elements $e_2 , e_3 , e_4$, we get the matrix
$[T]_{B}^{C} = \begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ -1 & 0 & -1 & 0 \\ 1 & 1 & -1 & -1 \end{bmatrix}.$

To find the rank of $T$, we row-reduce this matrix:
\begin{align*}
\begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ -1 & 0 & -1 & 0 \\ 1 & 1 & -1 & -1 \end{bmatrix} \xrightarrow[-R_3]{ \frac{1}{2} R_1 } \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & -1 & -1 \end{bmatrix} \xrightarrow{ R_3 – R_1 } \6pt] \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 1 & 1 & -1 & -1 \end{bmatrix} \xrightarrow{R_4 – R_1 – R_2 + R_3} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}. \end{align*} This reduced matrix has three non-zero rows, and so has rank 3. Thus the orginal linear transformation T also has rank 3. The nullity can be computed using the equation \[ \mathtt{rank}(T) + \mathtt{nullity}(T) = \dim (V). Using the values $\mathtt{rank}(T) = 3$ and $\dim (V) = 4$, the nullity of $T$ must be 1. Add to solve later

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