Determinant of a Block Lower Triangular Matrix

• Vespero
So you can use it to prove the result you want.In summary, the theorem states that the determinant of a block matrix with a k by k matrix A, and n by n matrix D, and n by k matrix C, is equal to the product of the determinants of A and D. This is proven by showing that the product of two block matrices, one with A and an identity matrix, and the other with C and D, is equal to the original block matrix. This is then further shown using the lemma which states that if all the entries in a row or column except for one entry vanish, then the determinant is equal to that entry multiplied by (-1)^(i+j) times the determinant of the matrix with the
Vespero

Homework Statement

Theorem. Let A be a k by k matrix, let D have size n by n and let C have size n by k. Then

$$det \left(\begin{array}{cc}A&0\\C&D\end{array}\right) = (det A)\cdot (det D)$$.

Proof. First show that
$$\left(\begin{array}{cc}A&0\\0&I_{n}\end{array}\right) \cdot \left(\begin{array}{cc}I_{k}&0\\C&D\end{array}\right) = \left(\begin{array}{cc}A&0\\C&D\end{array}\right)$$

Then use the following lemma:

Let A be an n by n matrix; let b denote its entry in row i and column j.
(a) If all the entries in row i other than b vanish, then
$$det A = b(-1)^{i+j}det A_{ij}$$

(b) The same equation holds if all the entries in column j other than the entry b vanish.

The Attempt at a Solution

I am attempting to show the first part of the problem, where our initial matrix is broken into two further matrices. Given the dimensions of the block matrices, or even writing out the entire matrices with individual entries and multiplying (since all multiplications are in essence the rows and columns of the block matrices multiplied either by standard base matrices or by zero matrices), it is readily apparent that the equation holds true. However, I am not sure how to write this out in a way that is concise, possibly utilizing sigma notation. (I haven't taken a "higher level" math class in a while, so even when things make sense intuitively or computationally, I'm not sure how to notate them.)

On the second part, (using the lemma), it is again apparent that the determinants of the two matrices simplify down to det A and det D by recursively using the lemma on the diagonal of the identity matrix until one is left with the determinant of the other matrix on the diagonal, but I'm not sure how to write this concisely.

In essence, I suppose my problem is less one of how to understand or solve the problem, but how to write my solution in a manner that isn't rambling and unelegant.

Vespero said:

Homework Statement

Theorem. Let A be a k by k matrix, let D have size n by n and let C have size n by k. Then

$$det \left(\begin{array}{cc}A&0\\C&D\end{array}\right) = (det A)\cdot (det D)$$.

Proof. First show that
$$\left(\begin{array}{cc}A&0\\0&I_{n}\end{array}\right) \cdot \left(\begin{array}{cc}I_{k}&0\\C&D\end{array}\right) = \left(\begin{array}{cc}A&0\\C&D\end{array}\right)$$

Then use the following lemma:

Let A be an n by n matrix; let b denote its entry in row i and column j.
(a) If all the entries in row i other than b vanish, then
$$det A = b(-1)^{i+j}det A_{ij}$$

(b) The same equation holds if all the entries in column j other than the entry b vanish.

The Attempt at a Solution

I am attempting to show the first part of the problem, where our initial matrix is broken into two further matrices. Given the dimensions of the block matrices, or even writing out the entire matrices with individual entries and multiplying (since all multiplications are in essence the rows and columns of the block matrices multiplied either by standard base matrices or by zero matrices), it is readily apparent that the equation holds true. However, I am not sure how to write this out in a way that is concise, possibly utilizing sigma notation. (I haven't taken a "higher level" math class in a while, so even when things make sense intuitively or computationally, I'm not sure how to notate them.)
I'd try writing it like this, though there might be a more elegant way to notate everything.

Write the product as M=QP where
$$Q_{ij} = \begin{cases} A_{ij} & \text{if } i\le k \text{ and } j \le k \\ 0 & \text{if } i\gt k \text{ and } j \le k \\ 0 & \text{if } i\le k \text{ and } j \gt k \\ \delta_{ij} & \text{if } i \gt k \text{ and } j \gt k \end{cases}$$ where ##\delta_{ij}## is the Kronecker delta. P is defined similarly (I'll leave it to you to do that). Then calculate ##M_{ij}## for the four cases separately. For example, when ##i\le k## and ##j \le k##, you have
$$M_{ij} = \sum_{l=1}^{n+k} Q_{il}P_{lj} = \sum_{l=1}^{k} Q_{il}P_{lj} + \sum_{l=k+1}^{n+k} Q_{il}P_{lj} = \sum_{l=1}^{k} A_{il}\delta_{lj} + \sum_{l=k+1}^{n+k} 0\times C_{(l-k)j} = A_{ij}$$

It might be neater to prove the more general result, that
$$\begin{pmatrix} A & B \\ C & D\end{pmatrix} \begin{pmatrix} P & Q \\ R & S\end{pmatrix} = \begin{pmatrix} AP + BR & AQ + BS \\ CP + DR & CQ + DS \end{pmatrix}$$
where the number of rows and columns in the partitioned matrices match up. Just write down the expression for a term in the product and split it into two parts.

That result is not restricted to square matrices, or some of the partitions being square.

What is a Block Lower Triangular Matrix?

A Block Lower Triangular Matrix is a type of square matrix where the elements above the main diagonal are all zero, and the remaining elements are further divided into smaller square blocks. These blocks are also lower triangular, meaning that the elements above the main diagonal within each block are also zero.

What is the determinant of a Block Lower Triangular Matrix?

The determinant of a Block Lower Triangular Matrix is the product of all the elements on the main diagonal. This means that the determinant of a Block Lower Triangular Matrix is calculated by multiplying the determinants of each individual block.

How is the determinant of a Block Lower Triangular Matrix calculated?

The determinant of a Block Lower Triangular Matrix is calculated by using the rule of expansion by minors. This involves choosing a row or column of the matrix, and then multiplying each element in that row or column by its corresponding minor (determinant of the submatrix formed by deleting the row and column of the element). These products are then summed to find the determinant.

What are the properties of the determinant of a Block Lower Triangular Matrix?

Some properties of the determinant of a Block Lower Triangular Matrix include: the determinant of the identity matrix is always 1, the determinant of a matrix and its transpose are equal, and the determinant of a matrix with two identical rows or columns is equal to 0.

What are some real-life applications of Block Lower Triangular Matrices?

Block Lower Triangular Matrices are commonly used in various fields of science, such as physics, engineering, and economics. They can be used to represent linear transformations, solve systems of equations, and analyze networks and interconnected systems. In economics, they can be used to model input-output relationships between different industries. In physics, they can be used to represent forces and interactions between objects.

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