Relating Associated Matrices and Changes of Basis

Topics: Algebra - Linear Transformation - Vector Space - Associated Matrix of a Linear Transformation

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Let $T:V\to W$ be a linear transformation.

(theorem)

Let $\beta$ and ${\beta }^{\prime }$ be ordered bases of $V$, and $\gamma$ and ${\gamma }^{\prime }$ ordered bases of $W$.

With the associated matrices of $T$ under these bases, we have that:

$$[T{]}_{{\beta }^{\prime }}^{{\gamma }^{\prime }}=P^{-1}[T{]}_{\beta }^{\gamma }Q$$

…where $Q$ is the change of basis matrix from ${\beta }^{\prime }$ to $\beta$, and $P$ the change of basis matrix that from ${\gamma }^{\prime }$ to $\gamma$.

Example

Let $T:{\mathbb{R}}^3\to {\mathbb{R}}^2$ be defined as $T(x,y,z)=(2x+y,x+y-z)$.

Let $\beta$ and $\gamma$ be the standard basis of ${\mathbb{R}}^3$ and ${\mathbb{R}}^2$, respectively.

Let:

• ${\beta }^{\prime }=\{(2,0,0),(0,-1,0),(0,0,-2)\}$ be a basis of ${\mathbb{R}}^3$
• ${\gamma }^{\prime }=\{(1,1),(1,-1)\}$ be a basis of ${\mathbb{R}}^2$

First step §

First, let’s calculate $[T{]}_{\beta }^{\gamma }$ and $[T{]}_{{\beta }^{\prime }}^{{\gamma }^{\prime }}$.

For $[T{]}_{\beta }^{\gamma }$, let’s represent the elements of $\beta$ as a linear combination of the elements of $\gamma$:

• $T(1,0,0)=(2,1)=2(1,0)+1(0,1)$
• $T(0,1,0)=(1,1)=1(1,0)+1(0,1)$
• $T(0,0,1)=(0,-1)=0(1,0)+(-1)(0,1)$
• With that:
  • $[T{]}_{\beta }^{\gamma }=\left(\begin{array}{ccc}2& 1& 0\\ 1& 1& -1\end{array}\right)$

For $[T{]}_{{\beta }^{\prime }}^{{\gamma }^{\prime }}$, let’s represent the elements of ${\beta }^{\prime }$ as a linear combination of the elements of ${\gamma }^{\prime }$:

• $T(2,0,0)=(4,2)=3(1,1)+1(1,-1)$
• $T(0,-1,0)=(-1,-1)=(-1)(1,1)+0(1,-1)$
• $T(0,0,-2)=(0,2)=1(1,1)+(-1)(1,-1)$
• With that:
  • $[T{]}_{{\beta }^{\prime }}^{{\gamma }^{\prime }}=\left(\begin{array}{ccc}3& -1& 1\\ 1& 0& -1\end{array}\right)$

Second step §

Second, let’s find $Q$.

For that, let’s see the elements of ${\beta }^{\prime }$ as a linear combination of the elements of $\beta$:

• $(2,0,0)=2(1,0,0)+0(0,1,0)+0(0,0,1)$
• $(0,-1,0)=0(1,0,0)+(-1)(0,1,0)+0(0,0,1)$
• $(0,0,-2)=0(1,0,0)+0(0,1,0)+(-2)(0,0,1)$
• With that:
  • $Q=\left(\begin{array}{ccc}2& 0& 0\\ 0& -1& 0\\ 0& 0& -2\end{array}\right)$

Third step §

Third, to find $P^{-1}$, let’s find $P$ first. Again, we do that by representing the elements of ${\gamma }^{\prime }$ as a linear combination of the elements of $\gamma$:

• (1, 1) = 1(1, 0) + 1(0, 1)$
• $(1,-1)=1(1,0)+(-1)(0,1)$
• With that:
  • $P=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$

Since $\det (P)=-2\ne 0$, we can invert $P$. We get that:

$P^{-1}=\left(\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right)$

Then we could multiply all of this to see that, in effect:

$$[T{]}_{{\beta }^{\prime }}^{{\gamma }^{\prime }}=P^{-1}[T{]}_{\beta }^{\gamma }Q$$

(corollary)

Let $T:V\to V$ be a linear transformation, with ordered bases $\beta$ and ${\beta }^{\prime }$. Then:

$$[T{]}_{{\beta }^{\prime }}=Q^{-1}[T{]}_{\beta }Q$$

…where $Q$ is the basis change matrix from ${\beta }^{\prime }$ to $\beta$.