Using a formula called Jacobi’s formula, we can deduce that for any square matrix ,
where is the matrix exponential discussed in homework 5. Suppose that a matrix is not invertible. Is invertible? Explain.
Based on the formula, we know that . No matter what is, . So , meaning that is invertible.
Suppose . Let and .
Can be a multiple of ? Explain your answer.
If for some scalar , then and . But then , which the problem says isn’t the case.
Let be a vector in . How many ways are there to write as a linear combination of and ? Explain your answer.
The number of ways to write as a linear combination of and is the same as the number of solutions to the equation,
Since the determinant of the matrix is not zero (as ), it is invertible. So, there is only one solution and hence exactly one way to write as a linear combination of and .
Let be the standard basis for the vector space . Suppose
is a linear transformation and that and .
Compute the matrix of using the basis .
The matrix . Plugging in we get that
Compute the trace of your matrix from (a).
If , then
is a basis for . Compute the matrix of using the basis .
The matrix , where is written as a column vector with respect to the ordered basis and is written as a column vector with respect to .
We can use our matrix from (a) to find out a few things. and (this is what it means to have a matrix for a linear transformation). How does one then write as a column vector with respect to ? Remember that
means that . In other words, that means that
is the solution to the equation , where
Since we know that is invertible, so . Similarly, . So,
This answer is fine, as is the answer where each entry is calculated out. But you might also notice that we can write a little more succinctly as the matrix product,
Compute the trace of your matrix from (c).
You can use an explicit matrix formula for (c) to find out the answer here. But you can also use properties of trace:
Find the value of for which
is in the set
Remember that asking if a vector is in a span is the same as asking if it can be written as a linear combination of the spanning vectors. But this is just a question about matrices! So, we rephrase the question in terms of an augmented matrix, and apply Gaussian elimination. We only need to get to row echelon form, since we don’t care about the actual solution, only “how many” there are.
For this augmented matrix to have a solution, the last row has to be only zeros. That is, we need that .
So the vector is in the span only when .