Linear Transformation, Basis For the Range, Rank, and Nullity, Not Injective

Linear Transformation problems and solutions

Problem 276

Let $V$ be the vector space of all $2\times 2$ real matrices and let $P_3$ be the vector space of all polynomials of degree $3$ or less with real coefficients.
Let $T: P_3 \to V$ be the linear transformation defined by
\[T(a_0+a_1x+a_2x^2+a_3x^3)=\begin{bmatrix}
a_0+a_2 & -a_0+a_3\\
a_1-a_2 & -a_1-a_3
\end{bmatrix}\] for any polynomial $a_0+a_1x+a_2x^2+a_3 \in P_3$.
Find a basis for the range of $T$, $\calR(T)$, and determine the rank of $T$, $\rk(T)$, and the nullity of $T$, $\nullity(T)$.
Also, prove that $T$ is not injective.

 
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Proof.

Let $\{1, x, x^2, x^3\}$ be a basis of the vector space $P_3$.
Then the range of $T$ is spanned by the images of these basis vectors. That is,
\[\calR(T)=\Span \{T(1), T(x), T(x^2), T(x^3)\}.\]

Thus, the range $\calR(T)$ is spanned by the following matrices
\begin{align*}
T(1)=\begin{bmatrix}
1 & -1\\
0& 0
\end{bmatrix}, T(x)=\begin{bmatrix}
0 & 0\\
1& -1
\end{bmatrix}\\[6pt] T(x^2)=\begin{bmatrix}
1 & 0\\
-1& 0
\end{bmatrix},
T(x^3)=\begin{bmatrix}
0 & 1\\
0& -1
\end{bmatrix}.
\end{align*}
We find a basis among these matrices.


Let $B=\{E_{11}, E_{12}, E_{21}, E_{22} \}$ be the standard basis of $V$, where the $(i,j)$-entry of $E_{ij}$ is $1$ and the other entries are $0$.
With respect to this basis $B$, the coordinate vectors of the above matrices are
\begin{align*}
[T(1)]_B=\begin{bmatrix}
1 \\
-1 \\
0 \\
0
\end{bmatrix},
[T(x)]_B=\begin{bmatrix}
0 \\
0 \\
1 \\
-1
\end{bmatrix}\\[6pt] [T(x^2)]_B=\begin{bmatrix}
1 \\
0 \\
-1 \\
0
\end{bmatrix},
[T(x^3)]_B=\begin{bmatrix}
0 \\
1 \\
0 \\
-1
\end{bmatrix}.
\end{align*}


Let $A$ be the matrix whose columns are these vectors:
\[A=\begin{bmatrix}
1 & 0 & 1 & 0 \\
-1 &0 & 0 & 1 \\
0 & 1 & -1 & 0 \\
0 & -1 & 0 & -1
\end{bmatrix}.\] Applying the elementary row operations, we obtain
\begin{align*}
A=\begin{bmatrix}
1 & 0 & 1 & 0 \\
-1 &0 & 0 & 1 \\
0 & 1 & -1 & 0 \\
0 & -1 & 0 & -1
\end{bmatrix}
\xrightarrow{\substack{R_2+R_1 \\ R_4+R_3}}
\begin{bmatrix}
1 & 0 & 1 & 0 \\
0 &0 & 1 & 1 \\
0 & 1 & -1 & 0 \\
0 & 0 & -1 & -1
\end{bmatrix}\\[10pt] \xrightarrow{\substack{R_2 \leftrightarrow R_3 }}
\begin{bmatrix}
1 & 0 & 1 & 0 \\
0 & 1 & -1 & 0 \\
0 &0 & 1 & 1 \\
0 & 0 & -1 & -1
\end{bmatrix}
\xrightarrow{\substack{R_1-R_3\\ R_2+R_3\\ R_4+R_3}}
\begin{bmatrix}
1 & 0 & 0 & -1 \\
0 & 1 & 0 & 1 \\
0 &0 & 1 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form and the first three columns contains the leading $1$’s.
Thus, the first three matrices $T(1), T(x), T(x^2)$ form a basis for the range $\calR(T)$.
(This is the “leading 1 method”.)

(Note also that the last column gives the coefficients of the linear combination $T(x^3)=-T(1)+T(x)+T(x^2)$.)


It follows that the rank, which is the dimension of the range, is $3$.

By the rank-nullity theorem we have
\[\rk(T)+\nullity(T)=\dim(P_3)=4.\] We deduce that the nullity is $1$.

Finally, since the nullity of $T$ is $1$, the linear transformation $T$ is not injective.
(Recall that $T$ is injective if and only if the nullity is $0$.)

Another way to see that $T$ is not injective is as follows.
We saw that $T(x^3)=-T(1)+T(x)+T(x^2)$. This implies by linearity that
\[T(x^3)=T(-1+x+x^2).\] Thus the values of $T$ at two distinct polynomials $x^3$ and $-1+x+x^2$ are the same.
Hence $T$ is not injective.


In conclusion,
\[\{T(1), T(x), T(x^2)\}=\left \{\begin{bmatrix}
1 & -1\\
0& 0
\end{bmatrix}, \begin{bmatrix}
0 & 0\\
1& -1
\end{bmatrix},
\begin{bmatrix}
1 & 0\\
-1& 0
\end{bmatrix}, \right \}
\] is a basis for the range of $T$. We have
\begin{align*}
\rk(T)=3 \text{ and } \nullity(T)=1.
\end{align*}


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