Answer:
First, it is easily shown that if the matrix is not square, then \({ A }^{ T }A\neq A{ A }^{ T }\) is generally not true because a when a matrix is transposed, it’s dimensions are changed.
\[\begin{bmatrix} { a }_{ 11 } & { a }_{ 12 } \\ { a }_{ 21 } & { a }_{ 22 } \\ { a }_{ 31 } & { a }_{ 32 } \end{bmatrix}*\begin{bmatrix} b_{ 11 } & { b }_{ 12 } & b_{ 13 } \\ b_{ 21 } & b_{ 22 } & b_{ 23 } \end{bmatrix}=\begin{bmatrix} c_{ 11 } & c_{ 12 } & c_{ 13 } \\ c_{ 21 } & c_{ 22 } & c_{ 23 } \\ c_{ 31 } & c_{ 32 } & c_{ 33 } \end{bmatrix}\]
Example:
A <- matrix(c(1, 2, 3, 4, 5, 6), nrow = 3)
mtx1 <- A %*% t(A)
mtx2 <- t(A) %*% A
all.equal(mtx1, mtx2)## [1] "Attributes: < Component \"dim\": Mean relative difference: 0.3333333 >"
## [2] "Numeric: lengths (9, 4) differ"For square matrices, it is generally not true because the vector multiplication of \({ A }^{ T }A\) would have to be the same as that for \(A{ A }^{ T }\). Let’s take the first calculation [1,1] of the resulting matrix for example.
A <- matrix(c(1, 2, 3, 4, 5, 6, 7, 8, 9), nrow = 3)
Arow1 = A[1, ]
ATransposeColumn1 = t(A)[, 1]
val1 = Arow1 %*% ATransposeColumn1
ATransposeRow1 = t(A)[1, ]
AColumn1 = A[, 1]
val2 = ATransposeRow1 %*% AColumn1
val1 == val2## [,1]
## [1,] FALSESo, for this to be true, the vector multiplciation results of both matrices would have to be equal, which means that \({ A }^{ T }A=A{ A }^{ T }\) would only work for a matrix which does not change when it is transpose is taken. Note, the proof is in the next answer to the question.
Answer:
As stated in the previous answer, for this statement to be true, \(A\) would have to be identical to \({ A }^{ T }\). A matrix with this property is called a symmetrical matrix.
A <- matrix(c(1, 2, 3, 2, 4, 2, 3, 2, 5), nrow = 3)
mtx1 <- A %*% t(A)
mtx2 <- t(A) %*% A
all.equal(mtx1, mtx2)## [1] TRUEGoing back to question 1.1, we can show the following:
For a symmetric matrix:
\[A={A}^{T}\] We can rewrite the equation provided to us in the assignment:
\[{ A }^{ T }A=A{ A }^{ T }\]
By substituting \({A}^{T}\) for \(A\), which is:
\[AA=AA\]
Here, we can see that order does not matter only because we are multiplying identical matrices on both sides of the equation.
Answer:
For this answer, we will use the matrix provided in the reading to test the function
lower_matrix <- function(A) {
mtx <- A
Imtx <- diag(dim(A)[1])
mtxs <- list()
i = 1
fixrow = 0
adj_mtx <- mtx
for (col in 1:as.integer(dim(mtx)[2])) {
for (row in 1:as.integer(dim(mtx)[1])) {
if (row == col & mtx[row, col] == 0) {
for (subrow in (row + 1):as.integer(dim(mtx)[1])) {
if (subrow > dim(mtx)[1]) {
stop("A more complex matrix solver is necessary")
}
if (mtx[subrow, col] != 0) {
tmprow = mtx[row, ]
mtx[row, ] = mtx[subrow, ]
mtx[subrow, ] = tmprow
fixrow = 1
adj_mtx <- mtx
}
}
if (fixrow == 0) {
stop("A more complex matrix solver is necessary")
}
}
if (row > col) {
Imtxtmp <- Imtx
Imtxtmp[row, col] <- -1 * mtx[row, col]/mtx[col, col]
mtxs[[i]] <- Imtxtmp
mtx <- Imtxtmp %*% mtx
i = i + 1
}
}
}
L <- solve(mtxs[[1]])
for (j in 2:(i - 1)) {
solve(mtxs[[j]])
L <- L %*% solve(mtxs[[j]])
}
return(list(A = adj_mtx, U = mtx, L = L))
}## [1] "A Matrix:"## [,1] [,2] [,3]
## [1,] 1 1 2
## [2,] 2 1 0
## [3,] 3 1 1## [1] "U Matrix"## [,1] [,2] [,3]
## [1,] 1 1 2
## [2,] 0 -1 -4
## [3,] 0 0 3## [1] "L Matrix"## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 2 1 0
## [3,] 3 2 1## [1] "Adjusted A Matrix"## [,1] [,2] [,3]
## [1,] 1 1 2
## [2,] 2 1 0
## [3,] 3 1 1## [1] "Result of LU, which should be equal to A_adj"## [,1] [,2] [,3]
## [1,] 1 1 2
## [2,] 2 1 0
## [3,] 3 1 1## [1] "Comparison of LU and A_adj"## [1] TRUE## [1] "A Matrix:"## [,1] [,2] [,3] [,4] [,5]
## [1,] 0 2 3 9 5
## [2,] 2 1 1 1 6
## [3,] 3 4 1 2 7
## [4,] 1 7 6 3 8
## [5,] 4 2 7 4 9## [1] "U Matrix"## [,1] [,2] [,3] [,4] [,5]
## [1,] 4 2 7.0 4.00 9.00000
## [2,] 0 2 3.0 9.00 5.00000
## [3,] 0 0 -2.5 -1.00 1.50000
## [4,] 0 0 0.0 -9.05 -10.80000
## [5,] 0 0 0.0 0.00 16.09392## [1] "L Matrix"## [,1] [,2] [,3] [,4] [,5]
## [1,] 1.00 0.00 0.0 0.000000 0
## [2,] 0.00 1.00 0.0 0.000000 0
## [3,] 0.50 0.00 1.0 0.000000 0
## [4,] 0.75 1.25 3.2 1.000000 0
## [5,] 0.25 3.25 2.2 2.767956 1## [1] "Adjusted A Matrix"## [,1] [,2] [,3] [,4] [,5]
## [1,] 4 2 7 4 9
## [2,] 0 2 3 9 5
## [3,] 2 1 1 1 6
## [4,] 3 4 1 2 7
## [5,] 1 7 6 3 8## [1] "Result of LU, which should be equal to A_adj"## [,1] [,2] [,3] [,4] [,5]
## [1,] 4 2 7 4 9
## [2,] 0 2 3 9 5
## [3,] 2 1 1 1 6
## [4,] 3 4 1 2 7
## [5,] 1 7 6 3 8## [1] "Comparison of LU and A_adj"## [1] TRUENote, because in some cases the matrix rows will be reorganized by my function, I compare \(LU\) to the adjusted matrix \(A\_adj\) which only varies form \(A\) in row order.