Eigenvalue Multiplicity: Finding K For Matrix A

Hey guys! Let's dive into a cool problem in linear algebra. We're going to figure out for which value of k the matrix A = [[4, k], [7, -8]] has one real eigenvalue with a multiplicity of 2. In simpler terms, we want to find the k that makes our matrix have a repeated eigenvalue. This is a classic problem that combines matrix operations with some algebraic thinking, and trust me, it’s super satisfying when you crack it. So, buckle up, and let's get started!

Understanding Eigenvalues and Multiplicity

Before we jump into the calculations, let's make sure we're all on the same page about eigenvalues and their multiplicity. Eigenvalues are special numbers associated with a matrix that tell us a lot about the matrix's behavior. Think of them as the matrix's fingerprints. Eigenvectors, on the other hand, are the vectors that, when multiplied by the matrix, only get scaled – they don't change direction. The eigenvalue is that scaling factor. The multiplicity of an eigenvalue is how many times it appears as a root of the characteristic equation. If an eigenvalue appears twice, we say it has a multiplicity of 2.

Now, why is this important? Well, when an eigenvalue has a multiplicity of 2, it means that the matrix has a repeated eigenvalue. For a 2x2 matrix, this has significant implications for the matrix's structure and behavior. Specifically, for our matrix to have one real eigenvalue of multiplicity 2, it means the characteristic equation must have a repeated root. This is the key concept we'll use to solve our problem. Understanding these basics is crucial because eigenvalues and eigenvectors are the cornerstones of many advanced topics in linear algebra, from diagonalizing matrices to solving systems of differential equations. So, grasping these concepts thoroughly will set you up for success in more complex mathematical landscapes.

Setting Up the Characteristic Equation

Alright, now that we've refreshed our understanding of eigenvalues, let's get our hands dirty with the math. The first step in finding the eigenvalues of a matrix is to set up the characteristic equation. The characteristic equation is given by det(A - λI) = 0, where A is our matrix, λ (lambda) is the eigenvalue, and I is the identity matrix. In our case, A = [[4, k], [7, -8]], so we need to subtract λI from A.

Let's break it down. The identity matrix I for a 2x2 matrix is [[1, 0], [0, 1]]. So, λI is [[λ, 0], [0, λ]]. Subtracting this from A, we get:

A - λI = [[4 - λ, k], [7, -8 - λ]]

Now, we need to find the determinant of this matrix. Remember, the determinant of a 2x2 matrix [[a, b], [c, d]] is ad - bc. So, the determinant of our matrix A - λI is:

(4 - λ)(-8 - λ) - (k)(7) = 0

Expanding this, we get:

λ² + 4λ - 32 - 7k = 0

This quadratic equation is our characteristic equation. The roots of this equation are the eigenvalues of our matrix A. And remember, we're looking for the value of k that makes this equation have a repeated root. This step is super important because the characteristic equation is the bridge that connects the matrix to its eigenvalues. By solving this equation, we're essentially unveiling the hidden properties of the matrix. So, make sure you're comfortable with setting up and simplifying this equation, as it's a fundamental skill in linear algebra.

Solving for k with Repeated Eigenvalues

Okay, we've arrived at the heart of the problem. We have our characteristic equation: λ² + 4λ - 32 - 7k = 0. Now, we need to find the value of k that makes this quadratic equation have a repeated root. Think back to your algebra days – a quadratic equation has a repeated root when its discriminant is equal to zero. The discriminant (Δ) of a quadratic equation ax² + bx + c = 0 is given by Δ = b² - 4ac.

In our equation, a = 1, b = 4, and c = -32 - 7k. So, the discriminant is:

Δ = 4² - 4(1)(-32 - 7k)

We want this discriminant to be zero for a repeated root, so:

16 + 4(32 + 7k) = 0

Simplifying, we get:

16 + 128 + 28k = 0

144 + 28k = 0

Now, let's solve for k:

28k = -144

k = -144 / 28

k = -36 / 7

So, the value of k that makes the matrix A have one real eigenvalue with a multiplicity of 2 is -36/7. This is a crucial step because setting the discriminant to zero is the key to finding the condition for repeated eigenvalues. Remember this trick – it's a lifesaver in many linear algebra problems! By ensuring the discriminant is zero, we're essentially forcing the quadratic equation to have a single, repeated solution, which corresponds to our matrix having a single eigenvalue with multiplicity 2. So, understanding and applying this concept is super important.

Verifying the Result

Awesome! We've found a value for k, but it's always a good idea to double-check our work. Let's plug k = -36/7 back into our characteristic equation and see if we indeed get a repeated root.

Our characteristic equation was: λ² + 4λ - 32 - 7k = 0

Substituting k = -36/7, we get:

λ² + 4λ - 32 - 7(-36/7) = 0

λ² + 4λ - 32 + 36 = 0

λ² + 4λ + 4 = 0

This is a perfect square trinomial! We can factor it as:

(λ + 2)² = 0

So, λ = -2 is indeed a repeated eigenvalue. This confirms that our value of k = -36/7 is correct! Yay! Verifying our results is a critical step in any mathematical problem-solving process. By plugging our solution back into the original equation, we ensure that our calculations are accurate and that our answer makes sense in the context of the problem. This step not only gives us confidence in our solution but also helps us catch any potential errors we might have made along the way. Remember, it's always better to be safe than sorry when it comes to math!

Conclusion

So, guys, we've successfully found the value of k that makes our matrix A have one real eigenvalue with a multiplicity of 2. We started by understanding the concepts of eigenvalues and their multiplicity, then we set up the characteristic equation, used the discriminant to find the condition for repeated roots, and finally, verified our result. This problem beautifully illustrates how different areas of math – linear algebra and quadratic equations – come together to solve a single problem.

Remember, the key takeaway here is the connection between the discriminant of the characteristic equation and the multiplicity of eigenvalues. This is a powerful concept that you can apply to many other problems in linear algebra. Keep practicing, and you'll become a pro at these types of problems in no time! And that's a wrap for today, folks. Keep your brains buzzing, and I'll catch you in the next math adventure!