Operations and Properties - Linear Algebra

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Question

True or False: If , are square and invertible matrices then is also invertible.

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Answer

To prove is invertible, we need to find another square matrix such that .

Since $A^{-1},B^{-1}$ exist, take $D = B^{-1}A^{-1}$ , then we have

$D(AB)=(B^{-1}A^{-1})AB = B^{-1}(A^{-1}A)B = B^{-1}B = I$ ,

and

$(AB)D = (AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AA^{-1} = I$ .

Hence is invertible.

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Question

Suppose that is an invertible matrix. Simplify $AAA^{-1}AAA^{-1}A^{-1}A$ .

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Answer

To simplify

$AAA^{-1}AAA^{-1}A^{-1}A$

we used the identities:

$AA^{-1}=A^{-1}A=I$

so we get

$AAA^{-1}AAA^{-1}A^{-1}A=A(AA^{-1})A(AA^{-1})(A^{-1}A)=AIAII=AA=A^2$

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Question

Suppose that are all invertible. What is the inverse of ?

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Answer

The inverse of is $C^{-1}B^{-1}A^{-1}$ since we can multiply it by to get:

$C^{-1}B^{-1}A^{-1}ABC=C^{-1}B^{-1}(A^{-1}A)BC=C^{-1}B^{-1}IBC$

$=C^{-1}B^{-1}BC=C^{-1}(B^{-1}B)C=C^{-1}C=I$

Therefore $C^{-1}B^{-1}A^{-1}$ is the inverse of

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Question

Determine the inverse of matrix A where

$A=\begin{bmatrix} 2& 3\ 4& 6 \end{bmatrix}$

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Answer

To determine the inverse of a matrix, you must first verify that the matrix is square. Next calculate the determinant. The determinant for this matrix is 0, so it does not have an inverse.

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Question

$\begin{align}&\text{Calculate the eigenvalues for the matrix }\&\begin{bmatrix}8&-2&-20\-2&-3&-9\-20&-9&-3\end{bmatrix}\end{align}$

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Answer

$\begin{align}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}8&-2&-20\-2&-3&-9\-20&-9&-3\end{bmatrix}\&\text{We can write a matrix of the form }\begin{bmatrix}8 - \lambda&-2&-20\-2&- \lambda - 3&-9\-20&-9&- \lambda - 3\end{bmatrix}\&\text{For a square matrix with dimensions of }3\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&\text{Knowing this, we can expand and solve:}\&(-1)\cdot (\lambda ^{3} - 2\lambda ^{2} - 524\lambda + 84)\&\lambda_{1}=-22.00;\lambda_{2}=0.16;\lambda_{3}=23.84\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align}$

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Question

and are both singular two-by-two matrices.

True or false: must also be singular.

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Answer

To prove a statement false, it suffices to find one case in which the statement does not hold. We show that

$A = \begin{bmatrix} 1& 0\ 0& 0 \end{bmatrix}$ and $B= \begin{bmatrix} 0& 0\ 0& 1 \end{bmatrix}$

provide a counterexample.

A matrix is singular - that is, without an inverse - if and only if its determinant is equal to zero. The determinant of a two-by-two matrix is equal to the product of its upper left to lower right entries minus that of its upper right to lower left entries, so:

$\det A = 1 \cdot 0 - 0 \cdot 0 = 0 - 0 = 0$

$\det B = 0 \cdot 1 - 0 \cdot 0 = 0 - 0 = 0$

Both and are singular.

Now add the matrices by adding them term by term.

$= \begin{bmatrix} 1& 0\ 0& 0 \end{bmatrix} + \begin{bmatrix} 0& 0\ 0& 1 \end{bmatrix}$

$= \begin{bmatrix} 1+0& 0+0\ 0+0 & 0+1 \end{bmatrix}$

$= \begin{bmatrix} 1 & 0\ 0 & 1 \end{bmatrix}$

This is simply the two-by-two identity, which has an inverse - namely, itself.

The statement has been proved false by counterexample.

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Question

and are both two-by-two matrices. has an inverse.

True or false: Both and have inverses.

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Answer

A matrix is nonsingular - that is, it has an inverse - if and only if its determinant is nonzero. Also, the determinant of the product of two matrices is equal to the product of their individual determinants. Combining these ideas:

$\det AB = \det A \cdot \det B$

If either $\det A = 0$ or $\det B = 0$ , then it must hold that

$\det AB = 0 \cdot 0 = 0$ .

Equivalently, if either or has no inverse, then has no inverse. Contrapositively, if has an inverse, it must hold that each of and has an inverse.

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Question

is an involutory matrix.

True, false, or indeterminate: 0 is an eigenvalue of .

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Answer

An eigenvalue of an involutory matrix must be either 1 or . This can be seen as follows:

Let $\lambda$ be an eigenvalue of involutory matrix . Then for some eigenvector $\textbf{v}$ ,

$A\textbf{v} = \lambda \textbf{v}$

Premultiply both sides by :

$AA\textbf{v} =A \lambda \textbf{v}$

$A^{2}\textbf{v} = \lambda A \textbf{v}$

By definition, an involutory matrix has as its square, so

$I \textbf{v} = \lambda A \textbf{v}$

$\textbf{v} = \lambda A \textbf{v}$

$\frac{1}{ \lambda }\textbf{v} =A \textbf{v}$

By transitivity,

$\lambda \textbf{v} = \frac{1}{ \lambda }\textbf{v}$

Thus, $\lambda = \frac{1}{\lambda}$ , or $\lambda^{2} = 1$

It follows that $\lambda = \pm 1$ . The statement is false.

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Question

and are both nonsingular two-by-two matrices.

True or false: must also be nonsingular.

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Answer

We can prove that the sum of two nonsingular matrices need not be nonsingular by counterexample.

Let $A = \begin{bmatrix} 1& 0\ 0& 1 \end{bmatrix}$ , $B= \begin{bmatrix} -1& 0\ 0& -1 \end{bmatrix}$ .

A matrix is nonsingular - that is, with an inverse - if and only if its determinant is nonzero. The determinant of a two-by-two matrix is equal to the product of its upper left to lower right entries minus that of its upper right to lower left entries, so:

$\det A = 1 \cdot 1 - 0 \cdot 0 = 1 -0 = 1$

$\det B = -1 \cdot (-1) - 0 \cdot 0 = 1 -0 = 1$

Both and are nonsingular.

Now add the matrices by adding them term by term.

$A + B= \begin{bmatrix} 1& 0\ 0& 1 \end{bmatrix}+ \begin{bmatrix} -1& 0\ 0& -1 \end{bmatrix}$

$= \begin{bmatrix} 1+ (-1)& 0+0 \ 0+0& 1+ (-1) \end{bmatrix}$

$= \begin{bmatrix} 0 & 0 \0 & 0 \end{bmatrix}$ ,

the zero matrix, whose determinant is 0 and which is therefore not nonsingular.

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Question

What is the dimension of the space spanned by the following vectors:

$\b{v}_1 = (1,1,0,0,0)$

$\b{v}_2 = (0,0,1,1,0)$

$\b{v}_3 = (0,0,0,0,1)$

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Answer

Since there are three linearly independent vectors, they span a 3 dimensional space.

Notice that the vectors each have 5 coordinates to them. Therefore they actually span a 3 dimensional subspace of a 5 dimensional space.

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Question

is a singular four-by-four matrix. True or false: must also be a singular matrix.

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Answer

A matrix is singular - that is, it has no inverse - if and only if its determinant is equal to 0. is singular, so

$\det R = 0$ .

The determinant of the scalar product of and an $n \times n$ matrix is

$\det (cR) = c^{n}(\det R)$ ;

setting , , $\det R = 0$ :

$\det (5R) = 5^{4 } \cdot 0$

$\det (5R) = 0$

Therefore, , having determinant 0, is also singular.

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Question

In a 5 dimensional vector space, what is the maximum number of vectors you can have in a linearly dependent set?

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Answer

Linearly dependent sets have no limit to the number of vectors they can have.

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Question

Consider the following set of three vectors:

$\b{v}_1,\b{v}_2,\b{v}_3$

where $\b{v}_3=2\b{v}_1+7\b{v}_2$

Is the set linearly independent?

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Answer

Since $\b{v}_3$ can be written as a linear combination of of $\b{v}_1$ and $\b{v}_2$ then the set cannot be linearly independent.

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Question

Does the following row reduced echelon form of a matrix represent a linearly independent set?

$\begin{bmatrix} 1 & 0& 0& 0& \0 &0 &1 &0 & \ 0 &0 &0 &0 &\end{bmatrix}$

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Answer

The set is linearly dependent because there is a row of all zeros.

Notice that having columns of all zeros does not tell if the set is linearly independent or not.

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Question

Does the following row reduced echelon form of a matrix represent a linearly independent set?

$\begin{bmatrix} 1 & 0& 0& 0& \0 &0 &1 &0 & \end{bmatrix}$

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Answer

The set must be linearly independent because there are no rows of all zeros. There are columns of all zeros, but columns do not tell us if the set is linearly independent or not.

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Question

is a nonsingular matrix.

True or false: the inverse of the matrix is $\frac{1}{3}R^{-1}$ .

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Answer

By definition,

$R R^{-1} = I$ and $R^{-1}R = I$ .

Multiply:

$3R \cdot \frac{1}{3}R^{-1} = 3 \cdot \frac{1}{3} \cdot RR^{-1} = 1 \cdot I = I$

Similarly,

$\frac{1}{3}R^{-1}\cdot 3R = \frac{1}{3} \cdot 3 \cdot R^{-1} R= 1 \cdot I = I$

Therefore, $\frac{1}{3}R^{-1}$ is the inverse of .

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Question

$A = \begin{bmatrix} 1 & 1 + i &0 \ 1 & 1 - i & 0 \ 0 & 1 & i \end{bmatrix}$

Find $A^{-1}$ .

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Answer

To find the inverse of a matrix , set up an augmented matrix , as shown below:

$\begin{bmatrix} 1 & 1 + i &0 &1 & 0 & 0 \ 1 & 1 - i & 0 & 0 & 1 & 0 \ 0 & 1 & i & 0 & 0 & 1 \end{bmatrix}$

Perform row operations on this matrix until it is in reduced row-echelon form.

The following operations are arguably the easiest:

$-R1+ R2 \rightarrow R2$

$\begin{bmatrix} 1 & 1 + i &0 &1 & 0 & 0 \ 0 & -2i & 0 & -1 & 1 & 0 \ 0 & 1 & i & 0 & 0 & 1 \end{bmatrix}$

$R2 \leftrightarrow R3$

$\begin{bmatrix} 1 & 1 + i &0 &1 & 0 & 0\ 0 & 1 & i & 0 & 0 & 1 \ 0 & -2i & 0 & -1 & 1 & 0 \end{bmatrix}$

$(-1 - i )R2 + R1 \rightarrow R1$

$(2i)R2 + R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0&1-i &1 & 0 & -1-i\ 0 & 1 & i & 0 & 0 & 1 \ 0 & 0 & -2 & -1 & 1 & 2i \end{bmatrix}$

$-\frac{1}{2} R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0&1-i &1 & 0 & -1-i\ 0 & 1 & i & 0 & 0 & 1 \ 0 & 0 &1 & \frac{1}{2} & - \frac{1}{2} & -i \end{bmatrix}$

$(-1+i)R3 + R1 \rightarrow R1$

$(- i)R3 + R2 \rightarrow R2$

$\begin{bmatrix} 1 & 0&0& \frac{1}{2}+\frac{1}{2}i & \frac{1}{2}-\frac{1}{2}i & 0\ 0 & 1 & 0& -\frac{1}{2} i & \frac{1}{2} i & 0 \ 0 & 0 &1 & \frac{1}{2} & - \frac{1}{2} & -i \end{bmatrix}$

The augmented matrix is in reduced row-echelon form $[I |A^{-1}]$ . The inverse is therefore

$A ^{-1} = \begin{bmatrix} \frac{1}{2}+\frac{1}{2}i & \frac{1}{2}-\frac{1}{2}i & 0\ -\frac{1}{2} i & \frac{1}{2} i & 0 \ \frac{1}{2} & - \frac{1}{2} & -i \end{bmatrix}$ .

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Question

Consider the mapping $f:R^2\rightarrow R^3$ . Can f be an isomorphism?

(Hint: Think about dimension's role in isomorphism)

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Answer

No, f, cannot be an isomorphism. This is because and have different dimension. Isomorphisms cannot exist between vector spaces of different dimension.

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Question

The last question showed us isomorphisms must be between vector spaces of the same dimension. This question now asks about homomorphisms.

Consider the mapping $f:R^2\rightarrow R^3$ . Can f be a homomorphism?

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Answer

The answer is yes. There is no restriction on dimension for homomorphism like there is for isomorphism. Therefore f could be a homomorphism, but it is not guaranteed.

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Question

In the previous question, we said an isomorphism cannot be between vector spaces of different dimension. But are all homomorphisms between vector spaces of the same dimension an isomorphism?

Consider the homomorphism $f:R^2\rightarrow R^2$ . Is f an isomorphism?

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Answer

The answer is not enough information. The reason is that it could be an isomorphism because it is between vector spaces of the same dimension, but that doesn't mean it is.

For example:

Consider the zero mapping f(x,y)= (0,0).

This mapping is not onto or 1-to-1 because all elements go to the zero vector. Therefore it is not an isomorphism even though it is a mapping between spaces with the same dimension.

Another example:

Consider the identity mapping f(x,y) = (x,y)

This is an isomorphism. It clearly preserves structure and is both onto and 1-to-1.

Thus f could be an isomorphism (example identity map) or it could NOT be an isomorphism ( Example the zero mapping)

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