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Linear Algebra Review

Differential Equations · Section 7.2 · generated practice set

Determinants, invertibility and independence — the six equivalent statements that eigenvalue work keeps using.

Practice this set → Fresh numbers on every attempt. No account needed.

Method

Only a few facts from linear algebra are needed, and they all orbit one question: is this matrix singular?

The equivalences

For a square matrix $A$, these say the same thing:

  • $det A eq 0$
  • $A$ is invertible
  • the columns of $A$ are linearly independent
  • $Amathbf{x} = mathbf{0}$ has only the trivial solution
  • $Amathbf{x} = mathbf{b}$ has exactly one solution for every $mathbf{b}$
  • $lambda = 0$ is not an eigenvalue of $A$

Negate all six and they still agree. In eigenvalue work the whole point of $det(A - lambda I) = 0$ is to find the $lambda$ that makes $A - lambda I$ singular, so that a non-zero eigenvector can exist.

Determinants

$$detegin{pmatrix} a & b \ c & dend{pmatrix} = ad - bc.$$ For $3 imes3$, expand along whichever row or column has the most zeros.

Linear independence of vector functions

For solutions of the same linear system, the Wronskian is the determinant of the matrix whose columns are the solution vectors: $W(t) = detleft[mathbf{x}^{(1)} mathbf{x}^{(2)} ight]$. As in 3.2, it is either identically zero or never zero, and non-zero means a fundamental set.

Definitions and theorems

Invertibility
$A$ is invertible iff $det A eq 0$, iff its columns are linearly independent, iff $Amathbf{x} = mathbf{0}$ has only the zero solution.
Wronskian of vector solutions
For solutions of $mathbf{x}' = A(t)mathbf{x}$, $W(t) = detleft[mathbf{x}^{(1)} cdots mathbf{x}^{(n)} ight]$ is either zero for all $t$ in the interval or zero for none.
Eigenvalue definition
$lambda$ is an eigenvalue of $A$ with eigenvector $mathbf{v} eq mathbf{0}$ iff $Amathbf{v} = lambdamathbf{v}$, equivalently $det(A - lambda I) = 0$.

Worked example

Are $mathbf{v}^{(1)} = egin{pmatrix} 1 \ 2end{pmatrix}$ and $mathbf{v}^{(2)} = egin{pmatrix} 3 \ 6end{pmatrix}$ linearly independent?

  1. Form the matrix with these as columns: $egin{pmatrix} 1 & 3 \ 2 & 6end{pmatrix}$.
  2. $det = (1)(6) - (3)(2) = 0$.
  3. A zero determinant means dependent — and indeed $mathbf{v}^{(2)} = 3mathbf{v}^{(1)}$.
  4. So they cannot form a fundamental set for a $2 imes2$ system: they span only a line, not the plane.
  5. Any eigenvector is only determined up to a scalar multiple, which is why two eigenvectors for the same eigenvalue are never automatically independent.

Common mistakes

  1. Putting the vectors in as rows and then worrying. $det A = det A^T$, so it makes no difference to independence — but for the Wronskian, stick to columns so the formula matches the textbook.
  2. Thinking a non-zero vector guarantees independence. Two non-zero vectors are dependent whenever one is a multiple of the other.
  3. Solving $(A - lambda I)mathbf{v} = mathbf{0}$ by row-reducing to the identity. If you can, $lambda$ is not an eigenvalue — the whole point is that the matrix is singular and the rows collapse.
  4. Expecting a unique eigenvector. Any non-zero multiple works. Fix a component to 1 to make the answer definite.

Practice it

Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.

Open 7.2 →

More sets in Differential Equations

1.1Introduction to ODEs & Direction Fields1.2Solutions to Some ODEs1.3Classification of Differential Equations2.1First-Order Linear Equations2.2Separable Equations2.3Modeling with First-Order ODEs