Change of Basis Matrix

Topics: Algebra - Vector Space - Basis

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Let $\beta$ and ${\beta }^{\prime }$ be ordered basis of $V$ a vector space on $\mathbb{K}$ with finite dimension.

Let $Q=[I_V{]}_{\beta }^{{\beta }^{\prime }}$ (i.e. the associated matrix to the identity transformation of $V$), then:

$$Q[v{]}_{\beta }=[v{]}_{{\beta }^{\prime }}\forall v\in V$$

That is, by multiplying $Q$ by the coordinate vector of $v\in V$ under $\beta$, we get the coordinate vector of $v$ under ${\beta }^{\prime }$.

We call $Q$ the change of basis matrix from $\beta$ to ${\beta }^{\prime }$. Note that $Q$ is invertible and that $Q^{-1}$ is the change of basis matrix from ${\beta }^{\prime }$ to $\beta$.

Example

Let $V={\mathbb{R}}^2$ and let:

• $\beta =\{(1,1),(1,-1)\}$
• ${\beta }^{\prime }=\{(2,4),(3,1)\}$

From here, we can represent the two elements of ${\beta }^{\prime }$ as a linear combination of the elements of $\beta$:

$$(2,4)=3(1,1)+(-1)(1,-1)$$

$$(3,1)=2(1,1)+1(1,-1)$$

Then, we can build a matrix with these scalars:

$$Q=\left(\begin{array}{cc}3& 2\\ -1& 1\end{array}\right)$$

This matrix can be seen as the matrix that represents elements of ${\beta }^{\prime }$ as elements of $\beta$.

Now, with the second proposition of this theorem, we have that:

$$[(2,4){]}_{\beta }=Q[(2,4){]}_{{\beta }^{\prime }}$$

So we can now easily get the coordinate vector of $v$ relative to $\beta$:

$$[(2,4){]}_{\beta }=\left(\begin{array}{cc}3& 2\\ -1& 1\end{array}\right)\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}3\\ -1\end{array}\right)$$