The rotation matrix \(R_\theta\) is a matrix whose multiplication rotates vectors in \(\mathbb R^2\) by \(\theta\) radians counterclockwise. \(R_\theta\) has the formula,
\[\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\]Calculate the determinant of \(R_\theta\). Does it depend on \(\theta\)?
The determinant is \(\cos^2(\theta)+\sin^2(\theta) = 1\), so it doesn’t depend on \(\theta\).
That the determinant is 1 means that the matrix \(R_\theta\) doesn’t change volumes of shapes.
Without doing any Gaussian elimination or using the formula for inverses of \(2 \times 2\) matrices, find a concise formula for \((R_\theta)^{-1}.\) (Hint: Think about what \(R_\theta\) does geometrically).
The matrix rotates everything simultaneously by \(\theta\) radians. How can you undo a rotation? Rotating the other way! You do that by rotating just the negative amount of radians.
So, \(R_\theta^{-1} = R_{-\theta}\)
Find a non-identity matrix \(B \ne I\) for which \(B^3 = BBB = I\).
We can use a rotation matrix for an example here. The matrix \(R_{2\pi/3}\) will rotate things by a third of the way around; rotating by this much exactly three times yields a full rotation. But nothing changes after a full rotation! So \(R_{2\pi/3}^3 = I\).
If \(A\) is an \(n \times n\) symmetric matrix (that is, \(A^T = A\)), and \(B\) is any \(n \times m\) matrix, show that the following are symmetric:
\(B^TB\)
We check, \((B^TB)^T = B^T (B^T)^T = B^TB\)
\(BB^T\)
We check, \((BB^T)^T = (B^T)^T B^T = BB^T\)
\(B^TAB\)
We check, \((B^TAB)^T = B^T A^T (B^T)^T = B^TAB\) because \(A^T = A\).
Find the following determinants:
The matrix is upper triangular, so the determinant is the product of its diagonals. So, the determinant is \((2)(-3)(-1)(-2) = -12\).
The first and last rows are multiples of each other. So, the matrix can’t be invertible, and its determinant is \(0\).
\(A\) is an \(n \times n\) matrix and \(\det(A) = 2\), find \(\det(3A)\). (Hint: First calculate \(\det(3I)\), then use that \(\det(AB) = \det(A)\det(B)\).)
\(3I_n\) is the \(n \times n\) matrix with 3s on its diagonal, so its determinant is the product of the diagonals. There are \(n\) copies of 3, so \(\det(3I)=3^n\).
The determinant is multiplicative, so \(\det(3A) = \det(3IA) = \det(3I)\det(A) = 3^n 2\).
Is \(\det\) a linear function on matrices?
No. Part (a) gave a counterexample to homogeneity (determinant doesn’t work well with scaling)
For an \(n \times n\) matrix \(A = (a^i_j)\) and a number \(\lambda\), compare \(\tr(\lambda A)\) and \(\lambda \tr(A)\).
They are the same, since sums are linear. So, trace works well with scaling.
For a pair of \(n \times n\) matrices \(A = (a^i_j)\) and \(B = (b^i_j)\), compare \(\tr(A+B)\) and \(\tr(A) + \tr(B)\).
They are the same, since sums are linear.
Is \(\tr\) a linear function on matrices?
Yes. We just proved the two things to check above.