Why Can't An Invertible Matrix Have An Eigenvalue Of 0?

An invertible matrix cannot have an eigenvalue of 0 because if it did, the matrix's determinant would be zero, making it non-invertible by definition. This fundamental relationship can be understood through several core concepts in linear algebra: the characteristic equation, the determinant-eigenvalue connection, the concept of the null space, and the properties of linear transformations.

1. The characteristic equation

A scalar λlambda

𝜆

is an eigenvalue of a matrix Acap A

𝐴

if it satisfies the characteristic equation:det(A−λI)=0det of open paren cap A minus lambda cap I close paren equals 0

det(𝐴−𝜆𝐼)=0

where Icap I

𝐼

is the identity matrix. The solutions to this polynomial equation are the eigenvalues of Acap A

𝐴

.

If we assume that 0 is an eigenvalue, we can substitute λ=0lambda equals 0

𝜆=0

into the characteristic equation:det(A−0I)=0det of open paren cap A minus 0 cap I close paren equals 0

det(𝐴−0𝐼)=0

det(A)=0det of open paren cap A close paren equals 0

det(𝐴)=0

This result, det(A)=0det of open paren cap A close paren equals 0

det(𝐴)=0

, is the condition for a matrix to be non-invertible, also known as a singular matrix. This creates a direct contradiction, proving that an invertible matrix cannot have an eigenvalue of 0.

2. Determinant and the product of eigenvalues

A critical property of matrices is that the determinant is equal to the product of all its eigenvalues.

det(A)=λ1×λ2×…×λndet of open paren cap A close paren equals lambda sub 1 cross lambda sub 2 cross … cross lambda sub n

det(𝐴)=𝜆1×𝜆2×…×𝜆𝑛

If even one eigenvalue, say λklambda sub k

𝜆𝑘

, is 0, then the product of all the eigenvalues must be 0:det(A)=λ1×λ2×…×0×…×λn=0det of open paren cap A close paren equals lambda sub 1 cross lambda sub 2 cross … cross 0 cross … cross lambda sub n equals 0

det(𝐴)=𝜆1×𝜆2×…×0×…×𝜆𝑛=0

As established earlier, a matrix with a determinant of 0 is not invertible. Therefore, if any eigenvalue is 0, the matrix cannot be invertible.

3. Null space and the invertible matrix theorem

The relationship can also be proven using the concept of a null space (or kernel), which is the set of all vectors that a matrix transforms into the zero vector. An eigenvalue of 0 is explicitly tied to the null space.

By definition, if 0 is an eigenvalue of matrix Acap A

𝐴

, there must exist a corresponding non-zero eigenvector xbold x

𝐱

such that:Ax=0x=0cap A bold x equals 0 bold x equals 0

𝐴𝐱=0𝐱=𝟎

This means that the non-zero vector xbold x

𝐱

lies in the null space of Acap A

𝐴

. A matrix is invertible if and only if its null space contains only the zero vector. The existence of a non-zero eigenvector xbold x

𝐱

in the null space violates this condition, proving that Acap A

𝐴

cannot be invertible.

4. Geometric interpretation

The geometric meaning of an eigenvalue of 0 is that the matrix transformation "crushes" or "squishes" space by collapsing at least one dimension into the zero vector.

• An invertible matrix represents a transformation that is both one-to-one (injective) and onto (surjective). It maps the vector space to itself without losing any dimensions.
• A matrix with a zero eigenvalue represents a transformation that is not one-to-one. It maps a non-zero vector (the eigenvector) to the zero vector. Since information is lost (the original vector cannot be recovered from the zero vector), the transformation cannot be inverted.

For example, a transformation that projects all vectors onto a plane will have a zero eigenvalue. Any vector perpendicular to that plane will be mapped to the zero vector, and there is no way to invert this process to retrieve the original vector.

Summary of relationships

The following table summarizes the equivalent properties that demonstrate why an invertible matrix cannot have an eigenvalue of 0.

Condition for Matrix Acap A 𝐴	Status	Reason
0 is an eigenvalue	Not invertible	There exists a non-zero vector xbold x 𝐱 such that Ax=0cap A bold x equals 0 𝐴𝐱=𝟎 .
det(A)=0det of open paren cap A close paren equals 0 det(𝐴)=0	Not invertible	The determinant is the product of the eigenvalues; if one eigenvalue is 0, the determinant is 0.
Non-trivial null space	Not invertible	A non-zero eigenvector for λ=0lambda equals 0 𝜆=0 exists in the null space.
Linearly dependent columns	Not invertible	The columns of the matrix are not a basis for the vector space.
Non-invertible transformation	Not invertible	The transformation collapses one or more dimensions, making it non-injective.