Problem 70 - Calculus for Biology and Medicine

Question

Let $$ A=\left[\begin{array}{rr} -1 & -2 \\ -4 & 1 \end{array}\right] $$ (a) Show that $$ \mathbf{u}_{1}=\left[\begin{array}{r} 1 \\ -2 \end{array}\right] \quad \text { and } \quad \mathbf{u}_{2}=\left[\begin{array}{l} 1 \\ 1 \end{array}\right] $$ are eigenvectors of \(A\) and that \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) are linearly independent. (b) Represent $$ \mathbf{x}=\left[\begin{array}{l} 1 \\ 2 \end{array}\right] $$ as a linear combination of \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\). (c) Use your results in (a) and (b) to compute \(A^{10} \mathbf{x}\).

Step-by-Step Solution

Answer

\( A^{10} \mathbf{x} = 3^9 \mathbf{x} \).

Step 1: Verify if \( \mathbf{u}_1 \) is an Eigenvector of \( A \)

To check if \( \mathbf{u}_1 \) is an eigenvector of \( A \), we compute \( A \mathbf{u}_1 \) and see if it is a scalar multiple of \( \mathbf{u}_1 \). Compute:\[A \mathbf{u}_1 = \left[\begin{array}{rr} -1 & -2 \ -4 & 1 \end{array}\right] \left[\begin{array}{r} 1 \ -2 \end{array}\right] = \left[\begin{array}{r} -1(1) + -2(-2) \ -4(1) + 1(-2) \end{array}\right] = \left[\begin{array}{r} 3 \ -6 \end{array}\right]\]Here, \( \left[\begin{array}{r} 3 \ -6 \end{array}\right] = 3 \mathbf{u}_1 \), showing \( \mathbf{u}_1 \) is an eigenvector with eigenvalue \( \lambda = 3 \).

Step 2: Verify if \( \mathbf{u}_2 \) is an Eigenvector of \( A \)

Next, check if \( \mathbf{u}_2 \) is an eigenvector of \( A \) in a similar way. Compute:\[A \mathbf{u}_2 = \left[\begin{array}{rr} -1 & -2 \ -4 & 1 \end{array}\right] \left[\begin{array}{l} 1 \ 1 \end{array}\right] = \left[\begin{array}{l} -1(1) + -2(1) \ -4(1) + 1(1) \end{array}\right] = \left[\begin{array}{l} -3 \ -3 \end{array}\right]\]Here, \( \left[\begin{array}{l} -3 \ -3 \end{array}\right] = -3 \mathbf{u}_2 \), indicating \( \mathbf{u}_2 \) is an eigenvector with eigenvalue \( \lambda = -3 \).

Step 3: Determine if \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) are Linearly Independent

To check linear independence, form a matrix \([ \mathbf{u}_1 \ \mathbf{u}_2 ]\) and compute its determinant. The vectors are linearly independent if the determinant is non-zero:\[\det\left( \left[ \begin{array}{cc} 1 & 1 \ -2 & 1 \end{array} \right] \right) = (1)(1) - (1)(-2) = 1 + 2 = 3\]Because the determinant is non-zero, \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) are linearly independent.

Step 4: Express \( \mathbf{x} \) as a Linear Combination of \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \)

We need coefficients \( c_1 \) and \( c_2 \) such that \( \mathbf{x} = c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 \). Form the equation:\[\left[ \begin{array}{l} 1 \ 2 \end{array} \right] = c_1 \left[ \begin{array}{r} 1 \ -2 \end{array} \right] + c_2 \left[ \begin{array}{l} 1 \ 1 \end{array} \right]\]Equate the components:1. \( c_1 + c_2 = 1 \)2. \( -2c_1 + c_2 = 2 \)Solve these simultaneous equations to find \( c_1 = \frac{1}{3} \) and \( c_2 = \frac{2}{3} \).

Step 5: Compute \( A^{10} \mathbf{x} \) Using Previous Results

Now, use the results to find \( A^{10} \mathbf{x} \):\[A^{10} \mathbf{x} = A^{10} \left( \frac{1}{3} \mathbf{u}_1 + \frac{2}{3} \mathbf{u}_2 \right) = \frac{1}{3} A^{10} \mathbf{u}_1 + \frac{2}{3} A^{10} \mathbf{u}_2\]Using the eigenvalues, we have:\[A^{10} \mathbf{u}_1 = 3^{10} \mathbf{u}_1\]\[A^{10} \mathbf{u}_2 = (-3)^{10} \mathbf{u}_2 = 3^{10} \mathbf{u}_2\]Thus, \( A^{10} \mathbf{x} = \frac{1}{3} \cdot 3^{10} \mathbf{u}_1 + \frac{2}{3} \cdot 3^{10} \mathbf{u}_2 = 3^9 (\frac{1}{3} \mathbf{u}_1 + \frac{2}{3} \mathbf{u}_2) = 3^9 \mathbf{x} \).

Step 6: Conclusion

As derived, the expression for \( A^{10} \mathbf{x} \) using the eigenvectors and their eigenvalues shows a scaled version of the original vector \( \mathbf{x} \):\[A^{10} \mathbf{x} = 3^9 \mathbf{x}\].

Key Concepts

Linear Combination

A linear combination is a way of combining vectors using scalar multiplication and addition. In essence, it’s what you do when you express one vector as a mixture of others. This is super useful in various mathematical and practical applications.
For example, if you have vectors \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \), a linear combination could look like: \( c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 \). Here, \( c_1 \) and \( c_2 \) are scalars. These can be any real numbers that adjust the magnitude of the vectors, effectively "mixing" them.

• When you express a vector as a linear combination of other vectors, it means you can replicate the vector using those basis vectors.
• This concept is crucial in understanding vector spaces and dimensions.

In the original exercise, \( \mathbf{x} = \left[ \begin{array}{l} 1 \ 2 \end{array} \right] \) is expressed as \( \frac{1}{3}\mathbf{u}_1 + \frac{2}{3}\mathbf{u}_2 \), which means \( \mathbf{x} \) can be reconstructed using \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) with coefficients \( \frac{1}{3} \) and \( \frac{2}{3} \), respectively.

Linear Independence

Linear independence is a critical concept that helps determine if a set of vectors are unique relative to each other. If vectors are linearly independent, no vector in the set can be described as a linear combination of the other vectors in that set.
This is important because:

• It defines the span of the vectors, informing us about the dimensions of the space they cover.
• Linearly independent vectors provide the maximum freedom of direction, valuable for basis vectors.

To check linear independence, we often use the determinant method. If the determinant of the matrix, formed by arranging vectors as its columns, is non-zero, the vectors are linearly independent. In contrast, a zero determinant indicates dependence.
In the solution provided, the vectors \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) are shown to be linearly independent because the determinant \( \det\left( \left[ \begin{array}{cc} 1 & 1 \ -2 & 1 \end{array} \right] \right) = 3 \) is non-zero. This confirms that no vector can be composed from one another using linear combinations.

Eigenvalues

Eigenvalues are a fundamental concept in linear algebra associated with eigenvectors. When a vector is acted upon by a matrix and the result is parallel to the original vector, this vector is called an eigenvector of the matrix, and the scalar by which it is stretched or shrunk is called an eigenvalue. Remember, eigenvalues and eigenvectors offer a powerful way to understand linear transformations.

When analyzing matrices, finding eigenvalues involves solving for \( \lambda \) in the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix.
In our exercise, we showed \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) are eigenvectors of matrix \( A \), with eigenvalues 3 and -3 respectively.

• \( \mathbf{u}_1 \) when multiplied by \( A \), equates to \( 3\mathbf{u}_1 \). This implies that the transformation scales \( \mathbf{u}_1 \) by the eigenvalue \( 3 \).
• Similarly, \( \mathbf{u}_2 \) is scaled by \(-3\), shown as \( A\mathbf{u}_2 = -3\mathbf{u}_2 \).

Understanding how eigenvalues relate to matrix transformations can help in simplifying complex systems.