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Linear Algebra : The Trace

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Example Questions

Example Question #21 : The Trace

$C = \begin{bmatrix} a& 0& -7& 3\\ 0& a& 2& -1\\ 0& 0&2 & 6\\ 0& 0& 0& 5 \end{bmatrix}$

where is a real number.

Which expression is equal to the trace of ?

Possible Answers:

2a+8

2a+7

$10a^{2}$

Correct answer:

2a+7

Explanation:

The trace of a matrix is equal to the sum of the elements in its main diagonal - the elements going from upper left to lower right. Therefore,

$\textrm{Tr}(A) = a+a+2+5 = 2a + 7$

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Example Question #22 : The Trace

$A = \begin{bmatrix} 2&0 &0 &0 \\ 0& x& 0& 0\\ 0& 0&6 &0 \\ 0&0 &0 & -4 \end{bmatrix}$ , $B= \begin{bmatrix} -4&0 &0 &0 \\ 0& 2& 0& 0\\ 0& 0&x &0 \\ 0&0 &0 & -3 \end{bmatrix}$

Give the trace of .

Possible Answers:

2x-1

-24x

8x+4

$x^{2}-x-20$

$-1,152x^{2}$

Correct answer:

8x+4

Explanation:

and are both diagonal matrices - their only nonzero elements are on their main diagonals, from upper left to lower right - so their product can be found by multiplying their corresponding diagonal elements:

$AB= \begin{bmatrix} 2&0 &0 &0 \\ 0& x& 0& 0\\ 0& 0&6 &0 \\ 0&0 &0 & -4 \end{bmatrix} \begin{bmatrix} -4&0 &0 &0 \\ 0& 2& 0& 0\\ 0& 0&x &0 \\ 0&0 &0 & -3 \end{bmatrix}$

$= \begin{bmatrix} 2(-4)&0 &0 &0 \\ 0& x(2)& 0& 0\\ 0& 0&6(x) &0 \\ 0&0 &0 & -4(-3) \end{bmatrix}$

$= \begin{bmatrix} -8&0 &0 &0 \\ 0&2x & 0& 0\\ 0& 0&6x&0 \\ 0&0 &0 & 12\end{bmatrix}$

The trace of a matrix is equal to the sum of the elements in its main diagonal, so the trace of is

$\textrm{Tr}(AB) = -8 + 2x + 6x + 12 = 8x+4$

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Example Question #21 : The Trace

is a square diagonal matrix, the trace of which is equal to 0.

True or false: must be singular.

Possible Answers:

True

False

Correct answer:

False

Explanation:

We can demonstrate this statement to be false with the counterexample

$A = \begin{bmatrix} 2&0 &0 \\ 0& -1& 0\\ 0&0 &-1 \end{bmatrix}$ .

This matrix fits the definition of a diagonal matrix in that it has zeroes everywhere except on its main (upper-left-to-lower-right) diagonal. Also, its trace - the sum of the elements on its main diagonal - is

Tr (A) = 2 + (-1 )+ (-1) = 0 .

A matrix is singular - that is, without an inverse - if and only if its determinant is equal to 0. The determinant of a diagonal matrix is equal to the product of its diagonal elements, so

$\det A = 2 (-1)(-1) = 2 \ne 0$ .

, a square diagonal matrix with trace 0, is not singular, proving the statement false through counterexample.

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Example Question #23 : The Trace

All skew-symmetric matrices have a trace of

Possible Answers:

It depends on the skew-symmetric matrix given

Correct answer:

Explanation:

For any skew-symmetric matrix , A^T = -A . Taking the trace of both sides we get

Tr(A)=Tr(-A)

Tr(A^T) = -Tr(A)

Tr(A) = -Tr(A)

2Tr(A)=0

Tr(A)=0 .

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Example Question #22 : The Trace

$A = \begin{bmatrix} 0& e& \pi \\ e & \pi & 1 \\ \pi & 2& e \end{bmatrix}$

Give the trace of .

Possible Answers:

$\pi + e$

$3\pi$

$\pi + e + 2$

$\pi + e + 1$

3 e

Correct answer:

$\pi + e$

Explanation:

The trace of the matrix is the sum of the elements on its main (upper-left corner to lower-right corner) diagonal:

$\textup{Tr}(A)= 0 + \pi + e = \pi + e$

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Example Question #23 : The Trace

is a diagonal matrix with trace 0.

True or false: It follows that $A ^{2}$ has trace 0.

Possible Answers:

False

True

Correct answer:

False

Explanation:

The statement is false, as can be demonstrated using the counterexample

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

This matrix is diagonal, so it can be squared by squaring its individual elements:

$A^{2} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

The trace of a matrix is equal to the sum of the elements on its main diagonal, so

$\textup{Tr }(A) = 1 + (-1 ) = 0$

but

$\textup{Tr }(A^{2}) = 1 +1 =2$

This gives us a matrix with trace 0 whose square does not have trace 0. The statement is false.

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Example Question #24 : The Trace

and are square matrices of the same dimension.

True or false: $\textup{Tr}(A) + \textup{Tr}(B) = \textup{Tr}(A+B)$

Possible Answers:

False

True

Correct answer:

True

Explanation:

The trace of a square matrix is equal to the sum of its diagonal elements - the elements whose row number and column number are equal. Therefore, if and are $N \times N$ matrices,

$\textup{Tr}(A) = \sum_{j=1}^ {N}a_{jj}$

and

$\textup{Tr}(B) = \sum_{j=1}^ {N}b_{jj}$

By the addition rule for finite sums,

$\textup{Tr}(A)+ \textup{Tr}(B) = \sum_{j=1}^ {N}a_{jj} + \sum_{j=1}^ {N}b_{jj} = \sum_{j=1}^ {N}(a_{jj} + b_{jj})$

Also, addition of matrices is elementwise, so

$\textup{Tr}(A + B) = \sum_{j=1}^ {N}(a_{jj} + b_{jj}) =\textup{Tr}(A)+ \textup{Tr}(B)$

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Example Question #28 : The Trace

The characteristic equation of a matrix is

$\lambda ^{4}+ 5\lambda ^{3}- 4 \lambda - 200 = 0$

Which statement must be true?

Possible Answers:

$\textup{Tr}(A) = 200$

$\textup{Tr}(A) = -5$

$\textup{Tr}(A) = -200$

None of the other statements are necessarily true.

$\textup{Tr}(A) = 5$

Correct answer:

$\textup{Tr}(A) = -5$

Explanation:

The sum of the eigenvalues of a matrix is equal to the trace of the matrix. If $\lambda_{1}, \lambda_{2}, \lambda_{3}, \lambda_{4}$ are the four (not necessarily distinct) eigenvalues of the matrix, then its characteristic equation is

$(\lambda - \lambda_{1}) (\lambda - \lambda_{2})(\lambda - \lambda_{3})(\lambda - \lambda_{4})= 0$

When expanded, the coefficient of its $\lambda^{3}$ term is $-( \lambda_{1}+ \lambda_{2}+ \lambda_{3}+ \lambda_{4})\lambda^{3}$ . It follows that the sum of the eigenvalues - and the trace of - is .

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Example Question #24 : The Trace

$B = \begin{bmatrix} 0 & 3 & -4 & 2 \\ 2 & 0 & 1 & -5 \\ -2 & 3 & 0 & 4 \\ -1 & 8 & 3 & 0 \end{bmatrix}$

True or false: The trace of is 0.

Possible Answers:

False

True

Correct answer:

True

Explanation:

The trace of a matrix is the sum of the elements along its main diagonal, which are shown below:

$B = \begin{bmatrix} \textbf{{\color{Red} 0}} & 3 & -4 & 2 \\ 2 & \textbf{{\color{Red} 0}} & 1 & -5 \\ -2 & 3 & \textbf{{\color{Red} 0}} & 4 \\ -1 & 8 & 3 & \textbf{{\color{Red} 0}} \end{bmatrix}$

The trace of can easily seen to be 0.

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Example Question #25 : The Trace

$B = \begin{bmatrix} i & 1 & -i & -1 \\ 1 & -i & -1& i \\ -i & -1& i & 1 \\-1& i & 1&-i \end{bmatrix}$

Give the trace of .

Possible Answers:

-4i

Correct answer:

Explanation:

The trace of a matrix is the sum of the elements along its main diagonal, which are shown below:

$B = \begin{bmatrix}\textbf{{\color{Red} i}} & 1 & -i & -1 \\ 1 & \textbf{{\color{Red} -i}} & -1& i \\ -i & -1& \textbf{{\color{Red} i}} & 1 \\-1& i & 1&\textbf{{\color{Red} -i}} \end{bmatrix}$

$\textup{Tr}(B) = i + (-i)+ i + (-i)= 0$

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Linear Algebra : The Trace

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #21 : The Trace

Example Question #22 : The Trace

Example Question #21 : The Trace

Example Question #23 : The Trace

Example Question #22 : The Trace

Example Question #23 : The Trace

Example Question #24 : The Trace

Example Question #28 : The Trace

Example Question #24 : The Trace

Example Question #25 : The Trace

All Linear Algebra Resources

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