I'm trying to work through how to calculate eigenvalues and eigenvectors.

I start with

$$Ax=\lambda x$$

Where $A$ is a $p \times p$ matrix, $\lambda$ is the eigenvalue and $x$ is the eigenvector.

This is the same as:

$$Ax=I\lambda x$$

$$Ax-I\lambda x=0$$

$$(A-I\lambda) x=0$$

We define the matrix $A$ as a $2 \times 2$ matrix:

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

Thus this -$I\lambda$ equals

$\begin{bmatrix}4-\lambda & -2\\-3 & 6-\lambda\end{bmatrix}$


$$Det(A-I\lambda)=24-10\lambda +\lambda^2 -6$$ $$Det(A-I\lambda)=18 - 10\lambda + \lambda^2 $$

Then, out of the blue my textbook claims that

$$0=30 - 10\lambda + \lambda^2 $$

How do I justify setting the determinant to $0$?

(I do "not" have an advanced knowledge in linear algebraic analysis, I only know how the determinant is used to calculate the inverse matrix)

  • 8
    $\begingroup$ Side note: is it $18-10\lambda+\lambda^2$ or $30-10\lambda+\lambda^2$? I think $18$ is correct, so there could be some typo in your book? $\endgroup$ – user491874 Jan 24 '18 at 12:35

The text is not claiming that the determinant is $0$. The text is saying "Let's find out for which values of lambda the determinant is $0$!"

So the determinant is $\lambda^2 - 10\lambda + 30$, and you want to find the $\lambda$ such that it is equal to zero. What do you do? You set it equal to zero and solve for $\lambda$. That is, you solve the equation

$$\lambda^2 - 10\lambda + 30 = 0$$

As for why you are interested in the values of $\lambda$ that make the determinant equal to $0$, remember that

$$rank(A-\lambda I) = n \iff det(A - \lambda I) \neq 0$$

So, if $det(A-\lambda I) \neq 0$, you will find that the only solution to $(A - \lambda I)x = 0$ is $x = 0$ (due to the fact that the rank of the matrix is full, hence the kernel only contains the $0$ vector). This means that the only $x$ such that $Ax = \lambda x$ is $x=0$, which means that $x$ is not an eigenvector.

So the only way to have eigenvectors is to have the determinant of $A - \lambda I$ be equal to zero, so that's why to find eigenvalues you look for the values of $\lambda$ that make $det(A - \lambda I) = 0$

| cite | improve this answer | |
  • $\begingroup$ I've got almost no prior knowledge of linear algebra (though we did cover matrices) so that formula is beyond me atm, I can't really see the connection to the explanation below. $\endgroup$ – Magnus Jan 24 '18 at 13:50
  • 7
    $\begingroup$ @Magnus I see. Well if you don't know that formula, though, it means that you can't really understand why to find the eigenvalues we set $det(A - \lambda I) = 0$. This step is crucial, otherwise you're left confused as why you're setting that determinant to zero. If you want to gain a deeper understanding you can start looking up the connection between number of solutions of a system, rank of a matrix, and determinant :) $\endgroup$ – Ant Jan 24 '18 at 14:09
  • 2
    $\begingroup$ The text is claiming that when it says that $A x = \lambda x$ and $x$ is an eigenvector. $\endgroup$ – JiK Jan 24 '18 at 16:13
  • 2
    $\begingroup$ @Magnus - you should wait at least 24 hours before accepting an answer, as a better one may come in. And an answer is not good if you don't understand it yourself. It doesn't matter if everyone else in the world understands it - if you don't, then it doesn't answer your question. You should interact with the asker and others to figure out the parts you don't follow. One you see what you are missing, then it is time to accept an answer (I don't say this in criticism of this answer - just some general principles.) $\endgroup$ – Paul Sinclair Jan 24 '18 at 17:33
  • 1
    $\begingroup$ @Magnus Rewording Paul's comment, you should actually accept an answer that is the most helpful for yourself regardless of the popularity (upvote or score). If this answer is the most helpful, then you can ignore this message. $\endgroup$ – Andrew T. Jan 25 '18 at 4:29

For a square matrix like $M = (A - \lambda I)$, the equation $Mx = 0$ will have a non-zero solution $x$ if and only if $M$ doesn't have an inverse, which is true if and only if the determinant of $M$ is $0$.

| cite | improve this answer | |
  • $\begingroup$ Does this imply that $Mx=0$ will only hold true if the determinant of $M$=0? $\endgroup$ – Magnus Jan 24 '18 at 12:31
  • $\begingroup$ @Magnus That's correct. That's also something you should already be familiar with from when you first covered determinants. $\endgroup$ – 5xum Jan 24 '18 at 12:32
  • $\begingroup$ Also, do this mean matrices can't have eigenvalues/eigenvectors if they have inverses? $\endgroup$ – Magnus Jan 24 '18 at 12:33
  • 4
    $\begingroup$ @Magnus $Mx = 0$ will always be true if $x$ is the $0$-vector. However, $Mx = 0$ will have a solution besides $x = 0$ when (and only when) the determinant of $M$ is $0$. A matrix will have an inverse if and only if $0$ is not one of its eigenvalues. $\endgroup$ – Ben Grossmann Jan 24 '18 at 12:34
  • 1
    $\begingroup$ @Magnus: You have your matrices mixed up. $A$ is the matrix we're trying to find eigenvalues for; it might have an inverse, or it might not. For any eigenvalue $\lambda$ of $A$, the matrix $M$ defined as $M=(A-\lambda I)$ definitely doesn't have an inverse. $\endgroup$ – user2357112 supports Monica Jan 24 '18 at 17:39

Here's a geometric interpretation of Ant's answer: the determinant tells you what happens to a unit volume of space after applying your transformation.

For example, the identity map $I$ leaves everything alone, so volume stays the same, so the determinant of $I$ is $1$. A multiple of the identity $rI$ stretches everything by a factor of $r$ in all $p$ directions, so the determinant of $rI$ is $r^p$.

In general, if your transformation $A$ has a set of eigenvectors $v_i$ which span your space then, in the direction of $v_i$, $A$ stretches things by a factor of the corresponding eigenvalue $λ_i$, and so overall it multiplies volume by the product of all the $λ_i$. So the determinant of $A$ is just the product of its eigenvalues - counted by multiplicity, i.e. according to how many independent eigenvectors each one has.

As for the property I stated, that the determinant equals the factor by which a volume of space increases in size: well, you can take that as a definition, and then check that it corresponds to the formula you're familiar with, e.g. by looking at what happens to the unit $p$-dimensional cube spanned by your basis vectors. (This definition also explains why it's involved in the calculation of inverses!)

To make the connection with your question explicit: if the determinant equals the product of the eigenvalues, then it will be zero exactly when one of them is zero. $A v = λ v$ is equivalent to $(A − λI) v = 0$, which says that $v$ is an eigenvector of $A − λ I$ with eigenvalue $0$, so the determinant of $A − λ I$ must be $0$ since it is the product of the eigenvalues.

| cite | improve this answer | |

The determinant of a $n\times n$ matrix $M$ is equal to $0$ if and only if the rank of the matrix is smaller than $n$, which happens if and only if the kernel of the matrix is non-empty, which happens if and only if there exists some vector $x\ne0$ such that $Mx=0$.

Therefore, $\lambda$ is an eigenvalue of $A$ $\iff$ the determinant of $A-\lambda I$ is equal to $0$.

| cite | improve this answer | |
  • $\begingroup$ I like this answer best, but perhaps falls outside of the OP's current working knowledge? $\endgroup$ – Kevin Jan 24 '18 at 12:35
  • 3
    $\begingroup$ @Kevin I think I only assume the knowledge that a student should have before covering the topic of eigenvalues. Most (if not all) linear algebra courses and textbooks introduce kernels, ranks, and determinants (and the properties that connect these concepts) before covering eigenvalues. I agree the OP might have been sloppy when covering that part, but in that case, my answer will be a good wake up call showing him he cannot just skip chapters in math and expect to not get lost. $\endgroup$ – 5xum Jan 24 '18 at 12:38
  • $\begingroup$ That's a very fair response. I agree with you on balance and further agree that the detail you covered should be known rather than simply following the mechanism of 'finding the $\lambda$'. $\endgroup$ – Kevin Jan 24 '18 at 12:55
  • $\begingroup$ @5xum While I do admit to having a lacking foundations, that's not so much a criticism of myself than of the course I'm taking. This being a course in statistics it doesn't require prior mathmatical knowledge "at all" (aside from high school mathmatics) and determinants, eigenvectors and eigenvalues were covered in something like 15 minutes. $\endgroup$ – Magnus Jan 24 '18 at 12:59
  • 1
    $\begingroup$ @Magnus In that case, yes, the course deserves some severe criticism. $\endgroup$ – 5xum Jan 24 '18 at 13:06

Let me give you some intuition that's not so algebraic but more geometric. Before I start, I'd like you to remember these:

  • A matrix represents some sort of linear transform from one space to another.
  • The determinant of a matrix represents how many times the volume of any object has after the transform, compared to its original volume.

You can think of the equation

$(A - \lambda I)x = 0$

as applying the transform $(A - \lambda I)$ to a vector $x$, so that $x$ becomes a zero vector.

No matter what that transform looks like, it must involve squashing the original space into a flatter space along the vector $x$, for example, squashing a 3D space into a 2D plane, or even a 1D line or a 0D point. The space after the transform is at least one dimension less than the original space, so the volume of any object in the original space becomes zero, so the determinant of $(A - \lambda I)$ has to be zero.

P.S. my intuition comes from this series of video given by 3Blue1Brown: Essence of Linear Algebra

| cite | improve this answer | |

I had the same question as well as initial response to the explanation by @Ant so maybe this might help.

I looked up the properties of invertible matrices https://en.wikipedia.org/wiki/Invertible_matrix#Properties and reasoned thus (changing the notation for the matrix from $A$ to $P$):

Recast $(P-I \lambda)$ $x$ $=$ $0$ as the familiar system of linear equations $Ax=b$.

If the matrix $A$ is invertible, then there exists exactly one solution to the equation $Ax=b$.

Since $b=0$, if $A$ is invertible then that one solution to the equation is $x=0$. But we don't want $x=0$ as that is not a useful result.

So let's make $A$ non-invertible. We can do that by setting $\det(A) = \det(P-I \lambda) = 0$.

| cite | improve this answer | |
  • $\begingroup$ Welcome to MSE. When you recast $(P-I\lambda)x=0$ as $Ax=b$, you already have $b=0$, so the unique solution is $x=0$ if $A$ is invertible and you don't need the "Since $b=0$.." line. The whole point is that we formulate the question so that the matrix $A=P-I\lambda$ is not invertible. This is so we can't write $A^{-1}Ax=x=0$, which is the trivial $x=0$ solution. Then we find the $\lambda$ that make such a thing true. $\endgroup$ – Daniel Buck Aug 16 '18 at 18:01

Note that if you try to find an eigenvector directly, and you take the coordinates of the eigenvector to be a,b then you have

$$Ax=\lambda x$$

$$A\begin{bmatrix}a \\b \end{bmatrix}=\lambda \begin{bmatrix}a \\b \end{bmatrix}$$

Expanding out that equation, you get

4a-2b = $\lambda $a

-3a+6b = $\lambda $b

which is equivalent to

(4-$\lambda $)a-2b = 0

-3a+(6-$\lambda $)b = 0

So b = (4-$\lambda $)a/2

Substituting that into the bottom equation

-3a+(6-$\lambda $)(4-$\lambda $)a/2 = 0

Factoring out a

-3+(6-$\lambda $)(4-$\lambda $)/2 = 0

-6+(6-$\lambda $)(4-$\lambda $) = 0

And now we're back to the equation that was derived from setting the determinant to zero. Setting the determinant of a matrix to zero is simply using the properties of matrices to get to that equation quicker.

| cite | improve this answer | |

Here is another way to look at your problem. You started with

$$Ax=I\lambda x$$

and you reasoned

$$Ax-I\lambda x=0$$

$$(A-I\lambda) x=0 \tag{1.}$$

Let the columns of $A-I\lambda$ be $v_1, v_2, \dots, v_n$. Then equation $(1.)$ can be written as

$$v_1x_1 + v_2x_2 + \cdots +v_nx_n = 0$$

In other words, the columns of $A-I\lambda$ are linearly dependent. That implies that the determinant of $A-I\lambda$ is zero.

| cite | improve this answer | |

As per the Eigen vectors for a transformation matrix , the transformation on such vector will result in scaling it to some values λ. Let X be the Eigen vector and A be the transformation matrix of X. Thus, AX = λX (λ being the scaling factor) AX-λX=0 (A-λ*I)*X = 0

Now this (A-λI) can be considered as another transformation matrix on X which needs to be 0. According to a algebraic rule if a transformation matrix transforms the vector X to zero , then its determinant must be zero. Which is exactly the condition we need to satisfy by finding λ(We call it Eigen values from here on) using det(A-λI)=0. Thus we make the condition det(A-λ*I) =0 and find λ (Eigen value) from it. Then supplying back Eigen values we get its corresponding Eigen vectors.

This is something I understood from why we need to use determinant .

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.