# Find Eigenvectors & Eigenvalues of An Matrix with All 1's Entries

• physicsss
In summary, to find n linearly independent eigenvectors of an n x n matrix with all 1's as its entries, we can use the characteristic equation det(A-\lambdaI)=0 and solve for n roots. The associated eigenvectors can be found by solving the equation (A-\lambdaI)v=0. For example, for the matrix A_{2x2}, the eigenvalues are \lambda=0,2 and the eigenvectors are v_1=[-1,1] and v_2=[0,1]. For the matrix A_{3x3}, the eigenvalues are \lambda=0,3 and the eigenvectors are v_1=[-1,1,0],

#### physicsss

A2 is a 2 x 2 matrix with all 1's as its entries, and A3 is a 3 x 3 matrix with all 1's as its entries, and An is an n x n matrix with all 1's as its entries. Find n linearly independent eigenvectors of An. What are their associated eigenvalues.

I have no idea how to do this. Any help would be super great!

EDIT:
Does v1=[1,0,...0], v2=[1,1,0,...0], vn=[1,1,1,...1] work? This is really hard since the null space of An only has n-1 linearly independent vectors.

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Find the determinant of each of those matrices and eventually you'll notice the relationship in a characteristic equation:

$$A = \left(\begin{array}{abcdef} 1-\lambda & 1 & 1\\ 1 & 1-\lambda & 1\\ 1 & 1 & 1-\lambda \end{array}\right)$$

$$det(A) = -\lambda^3+3\lambda^2=0$$

Similarly for $$A_{2x2}, det(A)=\lambda^2-2\lambda=0$$
$$A_{4x4}, det(A)=\lambda^4-4\lambda^3=0$$
$$A_{5x5}, det(A)=-\lambda^5+5\lambda^4=0$$

The $$\lambda$$ is the eigenvalue. You can find your associated eigenvector by solving for this equation:

$$Av=\lambda v$$

Where A is the matrix with all entries 1, v is the eigenvector, and $$\lambda$$ is the eigenvalue.

For example:

$$A_{2x2}:$$

$$A = \left(\begin{array}{abcdef} 1-\lambda & 1\\ 1 & 1-\lambda\\ \end{array}\right)$$

$$det(A) = (1-\lambda)(1-\lambda)-1=0$$

Solving for $$\lambda$$ we get: $$\lambda=0, 2$$

Now find associated eigenvectors for each eigenvalue:
$$Since Av=\lambda v$$
$$Av-\lambda v=0$$
$$(A-\lambda)v=0$$

So for $$\lambda=0:$$

$$A = \left(\begin{array}{abcdef} 1-0 & 1 \\ 1 & 1-0 \\ \end{array}\right)$$

$$A = \left(\begin{array}{abcdef} 1 & 1 \\ 1 & 1 \\ \end{array}\right)$$

Your eigenvector is of the form $$v = \left(\begin{array}{abcdef} v_1 \\ v_2 \\ \end{array}\right)$$

Multiplying out with vector v and equationg to 0 you get:
$$1v_1 + 1v_2 = 0$$
$$1v_1 + 1v_2 = 0$$

Therefore in this particular case $$1v_1 = -1v_2$$ you can pick any number for $$v_2$$, i'd go with $$v_2=1$$. So your eigenvector for $$\lambda=0$$ is:

$$v = \left(\begin{array}{abcdef} -1 \\ 1 \\ \end{array}\right)$$

You can check your eigenvalues and eigenvector simply by multiplying them out, since $$Av=\lambda v$$

Check:

$$\left(\begin{array}{abcdef} 1 & 1 \\ 1 & 1 \\ \end{array}\right)$$$$\left(\begin{array}{abcdef} -1 \\ 1 \\ \end{array}\right)$$$$=0*\left(\begin{array}{abcdef} -1\\ 1 \\ \end{array}\right)$$

However to answer your question for $$A_{nxn}$$ matrices..

since det(A)=0 for all such matrices and trace(A)=n you may notice that for every $$A_{nxn}$$ matrix you get n eigenvalues, but only one of them is an real number eigenvalue, $$\lambda=n$$. In other words, for $$A_{2x2}, \lambda=2,0; A_{3x3}, \lambda=3,0,0; A_{4x4}, \lambda=4,0,0,0$$ --- you get the idea.

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with no calculation at all, just thinking abut the meaning of eigenvalues, one immediately sees that there is an n-1 dimensional kernel, spanned by n-1 independent eigenvectors, namely e1-e2, e1-e3,...,e1-en.

then since every one of these standard vectors goes to (1,...,1), their sum which equals this vector, goes to (n,...,n), so e1+...+en is another eigenvector.

the moral here is to think about the meaning of things before calculating. this approach led me to see the answer almost instantly, even before getting as far as your EDIT, much less cronxeh's post.

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cronxeh said:
$$det(A) = (1-\lambda)(1-\lambda)-1=0$$

Solving for $$\lambda$$ we get: $$\lambda=0, 2$$

Now find associated eigenvectors for each eigenvalue:
$$Since Av=\lambda v$$
$$Av-\lambda v=0$$
$$(A-\lambda)v=0$$

Not to be a pain in the butt, but you forgot the idenity matrix (A-$$\lambda$$I)

## 1. What are eigenvectors and eigenvalues?

Eigenvectors and eigenvalues are important concepts in linear algebra. Eigenvectors are special vectors that do not change direction when multiplied by a certain square matrix. Eigenvalues are the corresponding scalar values that represent how much the eigenvectors are scaled by when multiplied by the matrix.

## 2. How do I find the eigenvectors and eigenvalues of a matrix with all 1's entries?

To find the eigenvectors and eigenvalues of a matrix with all 1's entries, you can use the following steps:

• Subtract the identity matrix from the matrix with all 1's entries.
• Find the determinant of the resulting matrix.
• The eigenvalues will be the roots of the characteristic polynomial.
• For each eigenvalue, solve the equation (A-I)x = 0 to find the corresponding eigenvector.

## 3. What is the significance of the eigenvectors and eigenvalues of a matrix?

Eigenvectors and eigenvalues have many important applications in science and engineering. They are used to solve systems of linear differential equations, analyze networks and circuits, and perform data analysis and image processing. They also have applications in quantum mechanics and statistics.

## 4. Can a matrix have more than one set of eigenvectors and eigenvalues?

Yes, a matrix can have multiple sets of eigenvectors and eigenvalues. This is because there are different ways to decompose a matrix into its eigenvectors and eigenvalues, and each decomposition may result in a different set of eigenvectors and eigenvalues.

## 5. Are there any special properties of a matrix with all 1's entries?

Yes, there are some special properties of a matrix with all 1's entries. For example, the sum of the rows and columns of such a matrix will always be equal to 1. Additionally, the eigenvalues of a matrix with all 1's entries will always add up to the number of rows or columns of the matrix.