Geometric Interpretation of Change of Basis

Standard Basis and Identity Matrix

Let $\vec{u}, \vec{v}, \vec{w}$ be the orthonormal vectors of the standard basis in $\mathbb{R}^3$

$$\begin{alignat*}{3} \vec{u} &= \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\\ \vec{v} &= \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\\ \vec{w} &= \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\\ \end{alignat*}$$

Let $\vec{p} = \begin{pmatrix} p_x \\ p_y \\ p_z \end{pmatrix}$ be an arbitrary vector in $\mathbb{R}^3$.

We can write $\vec{p}$ as the matrix-vector product $\boldsymbol{A} \vec{p} = \vec{p}$, where $\boldsymbol{A}$ is a $3 \times 3$ matrix whose rows represent the vectors $\vec{u}, \vec{v}, \vec{w}$:

$$\begin{alignat*}{3} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} p_x \\ p_y \\ p_z\end{pmatrix} = \begin{pmatrix} \vec{u} \cdot \vec{p} \\ \vec{v} \cdot \vec{p} \\ \vec{w} \cdot \vec{p} \end{pmatrix} \begin{pmatrix} p_x \\ p_y \\ p_z\end{pmatrix} \end{alignat*}$$

It is clear that the resulting components represent the scalar projections of $\vec{p}$ onto the axes of the standard basis.

These projections can be used to reconstruct $\vec{p}$ as a linear combination of the basis vectors:

Since

$$(\vec{u} \cdot \vec{p}) \vec{u} = (\frac{\vec{u} \cdot \vec{p}}{|\vec{u}|^2}) \vec{u} = (\frac{\vec{u} \cdot \vec{p}}{1}) \vec{u} = (p_x) \vec{u} = (p_x, 0, 0)^T$$

and Analogusly for $\vec{v}, \vec{w}$, we rewrite $\vec{p}$

$$p = (\vec{u} \cdot \vec{p}) \vec{u} + (\vec{v} \cdot \vec{p}) \vec{v} + (\vec{w} \cdot \vec{p}) \vec{w} = \begin{pmatrix}p_x \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix}0 \\ p_y \\ 0 \end{pmatrix} +\begin{pmatrix}0 \\ 0 \\ p_z \end{pmatrix} = \begin{pmatrix}p_x \\ p_y\\ p_z \end{pmatrix}$$

This construction generalizes to any orthonormal basis, as we show next.

Change of Basis with Orthonormal Axes

Claim: Let $\boldsymbol{A} \in \mathbb{R}^{3 \times 3}$ be a matrix whose rows are vectors $\vec{u}, \vec{v}, \vec{w}$ such that:

$$\begin{alignat*}{3} &|\vec{u}| = |\vec{v}| = |\vec{w}| = 1 \\ & \vec{u} \cdot \vec{v} = 0 \\ & \vec{u} \cdot \vec{w} = 0 \\ & \vec{v} \cdot \vec{w} = 0 \end{alignat*}$$

Then $\{ \vec{u}, \vec{v}, \vec{w} \}$ form an orthonormal basis of $\mathbb{R}^3$, and for any vector $\vec{p}$, the matrix-vector product

$$\boldsymbol{A}p = p'$$

returns the scalar projections of $\vec{p}$ onto the basis vectors.

Proof by Scalar Products:

Computing the product

$$\begin{alignat*}{3} \begin{pmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{pmatrix}\begin{pmatrix} p_x \\ p_y \\ p_z\end{pmatrix} \end{alignat*}$$

yields the vector

$$p' = \begin{pmatrix} u_x p_x + u_y p_y + u_z p_z \\ v_x p_x + v_y p_y + v_z p_z \\ w_x p_x + w_y p_y + w_z p_z \end{pmatrix} = \begin{pmatrix} \vec{u} \cdot \vec{p} \\ \vec{v} \cdot \vec{p} \\ \vec{w} \cdot \vec{p}\end{pmatrix}$$

Each component $\vec{b}_i \cdot \vec{p}$ is the scalar projection of $\vec{p}$ onto the basis vector $\vec{b}_i$.

Intuitively, the components in the resulting vector $p'$ show the scalar amount the component are showing into the respective direction.

To confirm that these projections are aligned with the original basis vectors, we rewrite:

$$\vec{p} = (\vec{u} \cdot \vec{p}) \vec{u} + (\vec{v} \cdot \vec{p}) \vec{v} + (\vec{w} \cdot \vec{p}) \vec{w}$$

Taking the dot product with any basis vector isolates its contribution:

$$\begin{alignat*}{3} p' \cdot \vec{u} &= (\vec{u} \cdot \vec{p})(\vec{u} \cdot \vec{u}) + (\vec{v} \cdot \vec{p})(\vec{v} \cdot \vec{u}) + (\vec{w} \cdot \vec{p})(\vec{w} \cdot \vec{u})\\ &= (\vec{u} \cdot \vec{p})(1) + (\vec{v} \cdot \vec{p})(0) + (\vec{w} \cdot \vec{p})(0) \\ & = \vec{u} \cdot \vec{p} \end{alignat*}$$

Analogously for $\vec{v}$ and $\vec{w}$.

Thus, the vector $\vec{p}' = \boldsymbol{A} \vec{p}$ contains exactly the scalar projections of $\vec{p}$ onto the axes of the new orthonormal basis, and the original vector can be reconstructed as a linear combination of the basis vectors scaled by those projections. $\Box$

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