Failure Example: A 2x2 Shear Matrix

Why a Simple Shear Can Fail to Diagonalize

A shear moves points in one direction by an amount that depends on their position in another direction. In two dimensions, the matrix can look almost like the identity: one coordinate changes by a multiple of the other, while the other coordinate stays fixed. That simple form makes a shear easy to compute with, but it also gives a useful warning: having a real eigenvalue does not guarantee that a matrix has enough independent eigenvectors to be diagonalized.

We work over \(\mathbb{R}\). A nonidentity horizontal shear has matrix \(\mathbf{S}_k=\begin{bmatrix}1&k\\0&1\end{bmatrix}\), where \(k\in\mathbb{R}\) and \(k\ne0\). It sends \((x,y)^{\top}\) to \((x+ky,y)^{\top}\). The parameter restriction matters: when \(k=0\), the matrix is the identity, not a nonidentity shear.

For the horizontal shear, the second coordinate is unchanged. That makes vectors on the \(x\)-axis natural candidates for eigenvectors. But to decide whether there are any other eigenvector directions, we must solve the full equation \(\mathbf{S}_k\mathbf{v}=\lambda\mathbf{v}\), not just test a convenient vector. The same calculation will explain why the shear cannot supply the two independent eigenvectors needed for a diagonalization.

The Eigenvectors of a Horizontal Shear

Write \(\mathbf{v}=(x,y)^{\top}\), with \((x,y)\ne(0,0)\). Multiplying the matrix by the vector and equating coordinates gives

The second equation forces the key case split. If \(y\ne0\), then \(1-\lambda=0\), so \(\lambda=1\). The first equation then becomes \(ky=0\), impossible because \(k\ne0\) and \(y\ne0\). Thus an eigenvector cannot have a nonzero second coordinate. If \(y=0\), the vector is nonzero only if \(x\ne0\); the first equation then forces \(\lambda=1\). Conversely, every vector \((x,0)^{\top}\) satisfies \(\mathbf{S}_k(x,0)^{\top}=(x,0)^{\top}\). Therefore the only eigenvalue is \(1\), and its eigenspace is exactly the \(x\)-axis.

“One independent eigenvector direction” does not mean that there is only one eigenvector. Every nonzero multiple of \((1,0)^{\top}\) is an eigenvector. The point is that any two such vectors are scalar multiples of each other, and therefore are linearly dependent. A diagonalization of a \(2\times2\) matrix requires two independent columns in its eigenvector matrix.

Proof. Suppose, for a contradiction, that \(\mathbf{S}_k=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), where \(\mathbf{P}\) is an invertible real \(2\times2\) matrix and \(\mathbf{D}\) is diagonal. Multiplying on the right by \(\mathbf{P}\) gives \(\mathbf{S}_k\mathbf{P}=\mathbf{P}\mathbf{D}\). Write the columns of \(\mathbf{P}\) as \(\mathbf{p}_1,\mathbf{p}_2\), and the diagonal entries of \(\mathbf{D}\) as \(d_1,d_2\). Comparing columns yields \(\mathbf{S}_k\mathbf{p}_1=d_1\mathbf{p}_1\) and \(\mathbf{S}_k\mathbf{p}_2=d_2\mathbf{p}_2\). Since \(\mathbf{P}\) is invertible, its columns are nonzero and linearly independent. They would therefore be two independent eigenvectors of \(\mathbf{S}_k\). But the eigenspace calculation shows that every eigenvector is a multiple of \((1,0)^{\top}\); any two such vectors are dependent. This contradiction proves that \(\mathbf{S}_k\) is not diagonalizable over \(\mathbb{R}\). \(\square\)

Worked Shear Matrices

What Changes When the Shear Is the Identity?

The condition that the shear be nonidentity is essential. For the family \(\mathbf{S}_k=\begin{bmatrix}1&k\\0&1\end{bmatrix}\), if \(k=0\), then \(\mathbf{S}_0=\mathbf{I}_2\). For every \((x,y)^{\top}\), the equation \(\mathbf{I}_2(x,y)^{\top}=1(x,y)^{\top}\) holds. In particular, \((1,0)^{\top}\) and \((0,1)^{\top}\) are independent eigenvectors, and \(\mathbf{I}_2=\mathbf{I}_2\mathbf{I}_2\mathbf{I}_2^{-1}\) is diagonalizable. When \(k\ne0\), the earlier calculation rules out every eigenvector with \(y\ne0\). So the zero-parameter case is not a small exception to overlook; it has a different eigenspace and a different diagonalization conclusion.

Common Mistake: Repeated Eigenvalue Does Not Mean Two Eigenvectors

A frequent error is to reason that because a \(2\times2\) shear has the eigenvalue \(1\) twice, it must have two independent eigenvectors. An eigenvalue equation does not work that way: the number of independent eigenvectors depends on the solutions of \((\mathbf{A}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}\), not just on how often the eigenvalue occurs in a polynomial.

For a concrete incorrect calculation, take \(\mathbf{A}=\begin{bmatrix}1&2\\0&1\end{bmatrix}\). A tempting but wrong claim is that \((0,1)^{\top}\) is a second eigenvector with eigenvalue \(1\), on the grounds that \(1\) is the only eigenvalue. Direct substitution corrects the claim:

Indeed, solving the equations gives \(2y=0\), hence \(y=0\), and then every eigenvector is a nonzero multiple of \((1,0)^{\top}\). The correction is to substitute a proposed eigenvector into \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\), and to solve the full equation before concluding that enough independent eigenvectors exist.

The key distinction is between an eigenvalue and its eigenspace. A nonidentity \(2\times2\) shear has a real eigenvalue, but all its eigenvectors lie on one line. Diagonalization requires two independent eigenvectors because the columns of the invertible matrix \(\mathbf{P}\) in \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) must be independent. The shear’s single eigenvector direction is therefore exactly what prevents diagonalization.