Three Checks for a Coordinate Conversion
A proposed change-of-basis matrix can look plausible and still convert coordinates in the wrong direction or contain an incorrect entry. Three checks help separate these issues: confirm the matrix dimensions, apply it to the coordinate vectors of the source basis, and compare the proposed reverse conversion by composing the matrices. Together, these checks test both what a matrix claims to do and whether the claimed reverse operation undoes it.
Work over \(\mathbb{R}\), and let \(V\) be a finite-dimensional vector space with ordered bases \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) and \(\mathcal{C}=(\mathbf{c}_1,\dots,\mathbf{c}_n)\). The coordinate vector \([\mathbf{v}]_{\mathcal{B}}\) is the column of coefficients in the unique expansion \(\mathbf{v}=x_1\mathbf{b}_1+\cdots+x_n\mathbf{b}_n\); \([\mathbf{v}]_{\mathcal{C}}\) is defined in the same way using \(\mathcal{C}\).
The arrow indicates the direction of conversion: coordinates in \(\mathcal{B}\) go in, and coordinates in \(\mathcal{C}\) come out. Both coordinate vectors have \(n\) entries, so this conversion matrix must be \(n\times n\). Checking dimensions can rule out an impossible proposal, but it cannot confirm that the entries or direction are correct.
Why the Columns Test the Conversion
Proof. First suppose \(\mathbf{A}\) gives the stated conversion for every vector. The \(j\)th standard coordinate vector \(\mathbf{e}_j\in\mathbb{R}^n\) is \([\mathbf{b}_j]_{\mathcal{B}}\), because \(\mathbf{b}_j\) has coefficient \(1\) on itself and \(0\) on the other vectors of \(\mathcal{B}\). Thus \(\mathbf{A}\mathbf{e}_j=[\mathbf{b}_j]_{\mathcal{C}}\). Multiplication by \(\mathbf{e}_j\) selects column \(j\) of \(\mathbf{A}\), proving the column condition. Conversely, suppose column \(j\) of \(\mathbf{A}\) is \([\mathbf{b}_j]_{\mathcal{C}}\) for every \(j\). Write any \(\mathbf{v}\in V\) as \(\mathbf{v}=\sum_{j=1}^{n}x_j\mathbf{b}_j\), so that \([\mathbf{v}]_{\mathcal{B}}=(x_1,\dots,x_n)^{\top}\). Then \(\mathbf{A}[\mathbf{v}]_{\mathcal{B}}=\sum_{j=1}^{n}x_j\mathbf{A}\mathbf{e}_j=\sum_{j=1}^{n}x_j[\mathbf{b}_j]_{\mathcal{C}}=[\mathbf{v}]_{\mathcal{C}}\). The last equality follows by expressing the same vector \(\mathbf{v}\) in the basis \(\mathcal{C}\). This proves the conversion formula for every \(\mathbf{v}\). \(\square\)
This theorem makes a basis-vector test decisive: a proposed matrix is correct precisely when each of its columns gives the right \(\mathcal{C}\)-coordinates for the corresponding \(\mathbf{b}_j\). Checking just one column can find an error, but passing one column does not verify the others.
Worked Example: Check the Dimensions and Both Basis Images
In \(\mathbb{R}^2\), let \(\mathcal{B}=((1,1)^{\top},(1,-1)^{\top})\) and \(\mathcal{C}=((1,0)^{\top},(1,1)^{\top})\). Check the proposed conversion \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}=\begin{bmatrix}0&2\\1&-1\end{bmatrix}\).
Each basis has two vectors, so a conversion from \(\mathcal{B}\)-coordinates to \(\mathcal{C}\)-coordinates must be \(2\times2\). The proposed matrix has that size. Now test its columns. To find \([(1,1)^{\top}]_{\mathcal{C}}\), solve \(a(1,0)^{\top}+b(1,1)^{\top}=(1,1)^{\top}\). Comparing coordinates gives \(a+b=1\) and \(b=1\), hence \(a=0\). For the second source vector, solve \(a(1,0)^{\top}+b(1,1)^{\top}=(1,-1)^{\top}\). This gives \(a+b=1\), \(b=-1\), so \(a=2\). Therefore the columns are \((0,1)^{\top}\) and \((2,-1)^{\top}\), exactly those in the proposal.
Check: The matrix sends \((1,0)^{\top}\) to \((0,1)^{\top}\), and \((1,1)^{\top}\) to \((2,-1)^{\top}\). These are the \(\mathcal{C}\)-coordinates of the two vectors in \(\mathcal{B}\), respectively, so the proposed matrix passes the basis-vector test.
Check the Reverse Conversion by Composition
The reverse conversion, from \(\mathcal{C}\)-coordinates back to \(\mathcal{B}\)-coordinates, is \(\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{C}}\). When the two conversions are correct, applying one after the other returns the starting coordinate vector. Thus the reverse matrix multiplied on the left of the forward matrix must give the identity. The opposite order must also give the identity because it starts with \(\mathcal{C}\)-coordinates instead.
Proof. Let \(\mathbf{x}=[\mathbf{v}]_{\mathcal{B}}\). By the definition of the forward conversion, \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}\mathbf{x}=[\mathbf{v}]_{\mathcal{C}}\). Converting back gives \(\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{C}}\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}\mathbf{x}=[\mathbf{v}]_{\mathcal{B}}=\mathbf{x}\). This holds for every \(\mathbf{x}\in\mathbb{R}^n\), since every such \(\mathbf{x}\) is the coordinate vector of some \(\mathbf{v}\in V\). In particular, taking \(\mathbf{x}=\mathbf{e}_j\) for each \(j\) shows that the product fixes every standard coordinate vector; its columns are therefore the columns of \(\mathbf{I}_n\), so the product equals \(\mathbf{I}_n\). Reversing the roles of \(\mathcal{B}\) and \(\mathcal{C}\) gives the other product in exactly the same way: starting from any \(\mathcal{C}\)-coordinate vector and converting to \(\mathcal{B}\) and back leaves it unchanged. Hence that product is also \(\mathbf{I}_n\). \(\square\)
These identities are a strong check on a proposed pair of matrices. However, matching products alone does not show that either matrix converts between the particular bases named: first check the basis-vector images, then check that the reverse conversions compose to identity.
Worked Example: Verify the Forward and Reverse Matrices
For the bases in the previous example, find and check \(\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{C}}\). To express \((1,0)^{\top}\) in \(\mathcal{B}\), solve \(a+b=1\), \(a-b=0\), giving \(a=b=\tfrac12\). To express \((1,1)^{\top}\) in \(\mathcal{B}\), solve \(a+b=1\), \(a-b=1\), giving \(a=1\), \(b=0\). These are the columns of the reverse matrix:
Multiply in both orders, showing the entries in each product:
Check: For example, start with \([\mathbf{v}]_{\mathcal{B}}=(3,-2)^{\top}\). The forward conversion gives \(\begin{bmatrix}0&2\\1&-1\end{bmatrix}(3,-2)^{\top}=(-4,5)^{\top}\). Converting back gives \(\begin{bmatrix}\tfrac12&1\\\tfrac12&0\end{bmatrix}(-4,5)^{\top}=(3,-2)^{\top}\), as required.
Using the Checks to Diagnose Errors
Worked Example: A Matrix with the Wrong Direction
Suppose someone labels \(\begin{bmatrix}\tfrac12&1\\\tfrac12&0\end{bmatrix}\) as \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}\) for the same bases. Its dimensions are \(2\times2\), so the dimension check does not detect a problem. Apply it to \([\mathbf{b}_1]_{\mathcal{B}}=(1,0)^{\top}\):
The output claims that \(\mathbf{b}_1\) has \(\mathcal{C}\)-coordinates \((\tfrac12,\tfrac12)^{\top}\). But these coordinates represent \(\tfrac12(1,0)^{\top}+\tfrac12(1,1)^{\top}=(1,\tfrac12)^{\top}\), not \(\mathbf{b}_1=(1,1)^{\top}\). The proposal is therefore not the stated conversion. Its columns are instead the \(\mathcal{B}\)-coordinates of the vectors in \(\mathcal{C}\), so it has the reverse direction.
Check: Use the correctly labelled matrices from the previous example. Their product in the order “reverse times proposed” is \(\begin{bmatrix}\tfrac12&1\\\tfrac12&0\end{bmatrix}\begin{bmatrix}\tfrac12&1\\\tfrac12&0\end{bmatrix}\) only if the mistaken label is treated as the forward matrix and the actual reverse is also used. More directly, composing the actual reverse with this mistaken forward gives \(\begin{bmatrix}\tfrac12&1\\\tfrac12&0\end{bmatrix}\begin{bmatrix}\tfrac12&1\\\tfrac12&0\end{bmatrix} =\begin{bmatrix}\tfrac34&\tfrac12\\\tfrac14&\tfrac12\end{bmatrix}\ne\mathbf{I}_2\). The failed round trip agrees with the basis-vector diagnosis.
Dimensions should be checked before multiplying: for a conversion from an \(n\)-vector coordinate column to another \(n\)-vector coordinate column, the matrix must have \(n\) rows and \(n\) columns. If a proposed change between two bases of the same space has, for instance, \(2\) rows and \(3\) columns, its output has two entries while the target basis coordinates require three. It cannot be the claimed conversion. A square matrix, however, may still have the wrong entries or the wrong direction, as the example shows.
Worked Example: Catch an Incorrect Column
Consider the proposed forward matrix \(\widetilde{\mathbf{P}}=\begin{bmatrix}0&2\\1&1\end{bmatrix}\) for the same bases. It has the right dimensions, and its first column passes the test because it is \([\mathbf{b}_1]_{\mathcal{C}}=(0,1)^{\top}\). Test the second column: it claims \([\mathbf{b}_2]_{\mathcal{C}}=(2,1)^{\top}\). In the basis \(\mathcal{C}\), this represents \(2(1,0)^{\top}+1(1,1)^{\top}=(3,1)^{\top}\), whereas \(\mathbf{b}_2=(1,-1)^{\top}\). Thus the second column is wrong; the correct coordinate column is \((2,-1)^{\top}\).
Composition with the correct reverse matrix provides a second diagnosis:
Check: The product is not \(\mathbf{I}_2\), so the proposed matrix does not round-trip correctly. Both tests identify the same error: the second column should be \((2,-1)^{\top}\), not \((2,1)^{\top}\).
A Reliable Audit
When checking a coordinate conversion, keep the source and target roles visible rather than relying on a matrix’s appearance. A useful order is:
- Identify the source basis and target basis from the arrow in the subscript.
- Check that the matrix dimensions take a source coordinate column to a target coordinate column.
- Apply the matrix to each standard coordinate vector. Confirm that the resulting column is the target-coordinate description of the corresponding source basis vector.
- If a reverse conversion is proposed, multiply in both orders and confirm that each product is the identity matrix of the appropriate coordinate space.
A common mistake is to check only that a matrix is square and then accept it as a change-of-basis matrix. For example, the incorrect matrix \(\begin{bmatrix}0&2\\1&1\end{bmatrix}\) above is square, but its second column represents \((3,1)^{\top}\), not the required source vector \((1,-1)^{\top}\). The correction is to test columns as basis-vector images, then check the round trip. The dimensions answer whether the proposal could have the required shape; the column and composition checks answer whether it actually does the claimed job.
Check Your Understanding
Use the source-to-target direction explicitly in each answer.
- If both bases have four vectors, what are the dimensions of the matrix converting coordinates from the first basis to the second?
- For a proposed matrix \(\mathbf{A}\) converting \(\mathcal{B}\)-coordinates to \(\mathcal{C}\)-coordinates, what must its \(j\)th column equal?
- Explain why a square matrix can pass the dimension check but still have the wrong direction.
- Write both composition identities for \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}\) and \(\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{C}}\).
- A proposed forward matrix has a second column that represents the wrong vector in the target basis. Which check reveals the error, and what should be done next?