Problem 9 - Differential Equations and Linear Algebra

Question

The effect of the linear transformation \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) with matrix \(A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\) is to reflect each vector across the line \(y=x .\) By arguing geometrically, determine all eigenvalues and eigenvectors of \(A\).

Step-by-Step Solution

Answer

In conclusion, the eigenvalues of the transformation matrix \(A\) are \(1\) and \(-1\). The eigenvectors corresponding to these eigenvalues are any vector of the form \( (v, v) \) for eigenvalue \(1\), and any vector of the form \( (v, -v) \) for eigenvalue \(-1\).

Step 1: Analyze the transformation

We're given that the transformation matrix \(A\) corresponds to a reflection across the line \(y=x\). So to find eigenvalues and eigenvectors, we need to determine which vectors are parallel to themselves after the reflection.

Step 2: Find vectors remaining parallel to themselves after reflection

Since we're reflecting the vectors across the line \(y=x\), the vectors that will remain parallel to themselves are those lying on the line \(y=x\), and those lying on the line \(y=-x\). Vectors of the form \( (v, v) \) lie on the line \(y=x\), and vectors of the form \( (v, -v) \) lie on the line \(y=-x\).

Step 3: Find eigenvalues

We have found two eigenvectors in Step 2, they lie on the lines \(y = x\) and \(y = -x\). Let's now determine their eigenvalues. To find eigenvalues, we need to check which scalar multiples of the original eigenvectors are reached after applying the transformation. For a vector lying on the line \(y=x\), the reflection will leave it unchanged, so the eigenvalue is \(1\). For a vector lying on the line \(y=-x\), the reflection will reverse its direction, so the eigenvalue is \(-1\).

Step 4: Find eigenvectors

Now that we have the eigenvalues, we need to find the eigenvectors. For eigenvalue \(1\), we're looking for vectors that lie on the line \(y=x\). One example of such a vector is \( (1,1) \), but any vector of the form \( (v, v) \) will work as an eigenvector. For eigenvalue \(-1\), we're looking for vectors that lie on the line \(y=-x\). One example of such a vector is \( (1,-1) \), and similarly, any vector of the form \( (v, -v) \) will work as an eigenvector.

Step 5: Conclusion

We have found that the eigenvalues of the transformation matrix \(A\) are \(1\) and \(-1\). The eigenvectors corresponding to these eigenvalues are any vector of the form \( (v, v) \) for eigenvalue \(1\), and any vector of the form \( (v, -v) \) for eigenvalue \(-1\).

Key Concepts

Linear Transformations

Understanding linear transformations is like unraveling the secret language of vector spaces. Picture a linear transformation as a function that takes a vector in one space and moves it to another, while keeping the gridlines straight and the origin fixed. This is akin to reshaping a pattern without tearing or crumpling it.

In our exercise, the linear transformation is represented by matrix A, which flips each vector across the line y=x. It's like looking in a mirror along the line y=x and seeing the reflection of your vector self! All the while, the rules of linearity—addition of vectors and multiplication by scalars—are respected, ensuring the transformation can be neatly described with just a matrix.

Reflection Across Line

When we talk about a reflection across a line in the context of linear transformations, envision a page reflected in water. This transformation flips the page over a specified line so that what was once on the left is now on the right, and vice versa.

Geometrically, you see two core types of lines in a reflection across y=x. First, vectors on y=x itself are like butterflies with symmetric wings—they stay the same. Second, vectors on y=-x are more rebellious: they flip to the opposite side, retaining the same angle but facing the other way. This neat geometric property helps us foresee which vectors will be our eigenvectors.

Eigenvalue Determination

Think of eigenvalues like secret spy codes that tell a vector exactly how it'll be stretched or squashed during the transformation. Determining eigenvalues is essentially asking, 'After transformation, by what factor has my vector been scaled?'

For a reflection, some vectors don't change their length but might flip direction. So, they either have an eigenvalue of 1 (same direction) or -1 (opposite direction). In our matrix A, the vectors on line y=x have an eigenvalue of 1 because they look the same post-reflection, while those on line y=-x get an eigenvalue of -1 since they're mirrored.

Eigenvector Determination

After we've cracked the code of the eigenvalues, we're on a treasure hunt for eigenvectors. An eigenvector says, 'Transform me, and I'll still point in the same line, just maybe backwards or stretched.'

In the search for eigenvectors of matrix A, we use the discovered eigenvalues. For the eigenvalue 1, we seek vectors on the line y=x. Any vector here is an eigenvector since it remains unchanged. Likewise, for the eigenvalue -1, vectors on y=-x serve as eigenvectors—flipping but staying on the same line. Remember, it's not about a single vector but a whole set of them that satisfy the condition for each eigenvalue.