From Preserving Lengths to Preserving Geometry
A linear map can change the coordinates of a vector while leaving its length unchanged. Such a map preserves more than individual lengths: it also preserves angles and dot products. The central question is whether preserving every norm forces this stronger geometric behavior, and how that behavior appears in the map’s matrix. Over \(\mathbb{R}^n\), the answer is precise: a linear map preserves norms exactly when its standard matrix is orthogonal.
Throughout, the dot product and norm on \(\mathbb{R}^n\) are the Euclidean ones, so \(\mathbf{u}\cdot\mathbf{v}=\mathbf{u}^{\top}\mathbf{v}\) and \(\|\mathbf{u}\|^2=\mathbf{u}\cdot\mathbf{u}\). The standard basis is \(\mathbf{e}_1,\dots,\mathbf{e}_n\). A linear map \(T:\mathbb{R}^n\to\mathbb{R}^n\) is determined by the images of these basis vectors: its standard matrix \(\mathbf{A}\) has columns \(T(\mathbf{e}_1),\dots,T(\mathbf{e}_n)\), and \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\).
The quantifier “for every” matters in the definition of an isometry. Preserving the lengths of a few selected vectors, even the standard basis vectors, does not by itself show that a map preserves all lengths. To understand what else is needed, first observe that the dot product can be recovered from norms.
Proof. Expand the first squared norm using the dot product: \(\|\mathbf{u}+\mathbf{v}\|^2=(\mathbf{u}+\mathbf{v})\cdot(\mathbf{u}+\mathbf{v})=\|\mathbf{u}\|^2+2\mathbf{u}\cdot\mathbf{v}+\|\mathbf{v}\|^2\). Subtract \(\|\mathbf{u}\|^2\) and \(\|\mathbf{v}\|^2\) from both sides, then divide by \(2\). \(\square\)
The Characterization
- \(T\) preserves norms: \(\|T(\mathbf{x})\|=\|\mathbf{x}\|\) for every \(\mathbf{x}\in\mathbb{R}^n\).
- \(T\) preserves dot products: \(T(\mathbf{u})\cdot T(\mathbf{v})=\mathbf{u}\cdot\mathbf{v}\) for every \(\mathbf{u},\mathbf{v}\in\mathbb{R}^n\).
- \(\mathbf{A}\) is orthogonal: \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{I}_n\).
Proof. Suppose first that \(T\) preserves norms. By linearity, \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\). Applying the polarization identity to \(T(\mathbf{u})\) and \(T(\mathbf{v})\) gives \(\displaystyle T(\mathbf{u})\cdot T(\mathbf{v})=\frac{\|T(\mathbf{u}+\mathbf{v})\|^2-\|T(\mathbf{u})\|^2-\|T(\mathbf{v})\|^2}{2}\). Norm preservation replaces the three squared norms in the numerator by \(\|\mathbf{u}+\mathbf{v}\|^2\), \(\|\mathbf{u}\|^2\), and \(\|\mathbf{v}\|^2\). The polarization identity then gives \(T(\mathbf{u})\cdot T(\mathbf{v})=\mathbf{u}\cdot\mathbf{v}\). This proves (1) implies (2).
Suppose next that \(T\) preserves dot products. Taking \(\mathbf{v}=\mathbf{u}=\mathbf{x}\) gives \(\|T(\mathbf{x})\|^2=T(\mathbf{x})\cdot T(\mathbf{x})=\mathbf{x}\cdot\mathbf{x}=\|\mathbf{x}\|^2\). Both norms are nonnegative, so equality of their squares gives equality of the norms. Thus (2) implies (1).
Write the columns of \(\mathbf{A}\) as \(\mathbf{a}_j=T(\mathbf{e}_j)\). The entry in row \(i\), column \(j\) of \(\mathbf{A}^{\top}\mathbf{A}\) is \(\mathbf{a}_i\cdot\mathbf{a}_j\). If \(T\) preserves dot products, this is \(T(\mathbf{e}_i)\cdot T(\mathbf{e}_j)=\mathbf{e}_i\cdot\mathbf{e}_j\), which is \(1\) when \(i=j\) and \(0\) otherwise. Hence \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{I}_n\), proving (2) implies (3).
Finally, suppose \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{I}_n\). Since \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\), \(\|T(\mathbf{x})\|^2=(\mathbf{A}\mathbf{x})^{\top}(\mathbf{A}\mathbf{x})=\mathbf{x}^{\top}\mathbf{A}^{\top}\mathbf{A}\mathbf{x}=\mathbf{x}^{\top}\mathbf{x}=\|\mathbf{x}\|^2\). Again, nonnegativity of norms gives \(\|T(\mathbf{x})\|=\|\mathbf{x}\|\). Thus (3) implies (1), and all three conditions are equivalent. \(\square\)
The proof identifies the crucial role of linearity: it lets us express the image of a sum as the sum of the images. Norm preservation can then be applied to \(\mathbf{u}+\mathbf{v}\), not just to \(\mathbf{u}\) and \(\mathbf{v}\) separately. The resulting dot-product preservation says that the columns of the standard matrix are unit vectors and are mutually orthogonal.
Worked Applications
Worked Example: Verify a Linear Isometry from Its Matrix
Consider the linear map \(T:\mathbb{R}^2\to\mathbb{R}^2\) with standard matrix \(\displaystyle \mathbf{A}=\frac{1}{5}\begin{bmatrix}3&-4\\4&3\end{bmatrix}\). Decide whether \(T\) is a linear isometry, then check its action on \(\mathbf{x}=(1,2)^{\top}\).
The columns are \(\mathbf{a}_1=(3/5,4/5)^{\top}\) and \(\mathbf{a}_2=(-4/5,3/5)^{\top}\). Their dot products are \(\mathbf{a}_1\cdot\mathbf{a}_1=9/25+16/25=1\), \(\mathbf{a}_2\cdot\mathbf{a}_2=16/25+9/25=1\), and \(\mathbf{a}_1\cdot\mathbf{a}_2=-12/25+12/25=0\). Therefore \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{I}_2\), so the theorem says \(T\) preserves every norm.
For the requested direct check, multiply: \(\displaystyle T(\mathbf{x})=\mathbf{A}\mathbf{x}=\frac{1}{5}\begin{bmatrix}3&-4\\4&3\end{bmatrix}\begin{bmatrix}1\\2\end{bmatrix}=\begin{bmatrix}-1\\2\end{bmatrix}\). Thus \(\|\mathbf{x}\|^2=1^2+2^2=5\) and \(\|T(\mathbf{x})\|^2=(-1)^2+2^2=5\). The computed image has the same norm, as the theorem predicts.
Worked Example: Find All Parameters That Give an Isometry
For real \(a,b\), define \(T(x,y)=(x,ax+by)\). Find all pairs \((a,b)\) for which \(T\) preserves norms.
For an arbitrary input \((x,y)^{\top}\), the squared output norm is \(\|T(x,y)\|^2=x^2+(ax+by)^2=(1+a^2)x^2+2abxy+b^2y^2\). The input squared norm is \(x^2+y^2\). If the norms are equal for every \(x,y\), set \(y=0\) and \(x=1\). The equality becomes \(1+a^2=1\), so \(a^2=0\), hence \(a=0\). Set \(x=0\) and \(y=1\); equality gives \(b^2=1\), so \(b=1\) or \(b=-1\).
Both possibilities work: when \(a=0\) and \(b=\pm1\), the map is \(T(x,y)=(x,\pm y)\), and \(\|T(x,y)\|^2=x^2+(\pm y)^2=x^2+y^2=\|(x,y)\|^2\) for every \(x,y\). Its standard matrix is \(\begin{bmatrix}1&0\\0&b\end{bmatrix}\), and for either permitted value of \(b\), its transpose times itself is \(\begin{bmatrix}1&0\\0&b^2\end{bmatrix}=\mathbf{I}_2\). The full answer is \(a=0\), \(b=\pm1\).
Worked Example: Why Unit-Length Columns Are Not Enough
Consider \(\displaystyle \mathbf{A}=\begin{bmatrix}1&3/5\\0&4/5\end{bmatrix}\). Both columns have unit length: the first has squared length \(1\), and the second has squared length \(9/25+16/25=1\). Does the corresponding linear map preserve every norm?
The columns are not orthogonal, since their dot product is \(3/5\), not \(0\). Equivalently, \(\displaystyle \mathbf{A}^{\top}\mathbf{A}=\begin{bmatrix}1&3/5\\3/5&1\end{bmatrix}\ne\mathbf{I}_2\). By the theorem, this matrix does not define a linear isometry. A direct check exhibits the failure. Take \(\mathbf{x}=(1,1)^{\top}\). Then \(\displaystyle \mathbf{A}\mathbf{x}=\begin{bmatrix}1+3/5\\4/5\end{bmatrix}=\begin{bmatrix}8/5\\4/5\end{bmatrix}\). The input has squared norm \(1+1=2\), while the output has squared norm \((8/5)^2+(4/5)^2=64/25+16/25=80/25=16/5\). Since \(16/5\ne2\), this input’s norm is not preserved.
Check: the two columns do have length \(1\), but their nonzero dot product prevents the matrix from being orthogonal. Checking column lengths alone is not enough.
Common Pitfall: Testing Only the Basis Vectors
A common incorrect argument is: “The images of the standard basis vectors have length \(1\), so the map preserves all lengths.” The matrix in the preceding example shows why this calculation fails. It maps \(\mathbf{e}_1\) to \((1,0)^{\top}\) and \(\mathbf{e}_2\) to \((3/5,4/5)^{\top}\), each of length \(1\). But it does not preserve the length of \(\mathbf{e}_1+\mathbf{e}_2\). The missing condition is that the images must also be mutually orthogonal.
For a square standard matrix, the complete test is \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{I}_n\): this simultaneously checks that every column has length \(1\) and that every pair of distinct columns has dot product \(0\). Alternatively, one can use the theorem’s norm criterion directly, but it must hold for every vector, not just for selected inputs. The polarization identity explains why the universal condition is strong enough: if all lengths are preserved, then the lengths of sums are preserved too, and that recovers every dot product.
Why the Characterization Is Useful
The theorem connects three ways of describing the same geometry. Norm preservation is a statement about the action of a linear map on all vectors; dot-product preservation describes the relationships between every pair of images; orthogonality is a finite matrix test on the standard matrix. In calculations, the matrix test is often the most efficient. In proofs about lengths, angles, or orthogonality, the map formulation may be more natural.
For example, if a map is known to be linear and its standard matrix has orthonormal columns, the theorem immediately guarantees that every vector keeps its length. Conversely, if a linear map is specified through a geometric property of preserving all norms, the polarization identity lets us infer that it preserves dot products and that its standard matrix is orthogonal. These implications rely on the stated assumptions: the map is linear, it maps \(\mathbb{R}^n\) to itself, and norm preservation is required for every input.
Check Your Understanding
Use the equivalence and the polarization identity to justify your answers.
- State the polarization identity for \(\mathbf{u},\mathbf{v}\in\mathbb{R}^n\).
- Why does linearity matter when using norm preservation to recover dot products?
- A linear map has a standard matrix whose columns all have length \(1\). What additional condition is needed for the matrix to be orthogonal?
- For which real \(c\) does \(T(x,y)=(cx,y)\) preserve norms for every \((x,y)\in\mathbb{R}^2\)?
- If a linear map preserves dot products, explain why it must preserve norms.
