Is there a fast way to prove a tridiagonal matrix is positive definite?

I' m trying to prove that

$$A=\begin{pmatrix} 4 & 2 & 0 & 0 & 0 \\ 2 & 5 & 2 & 0 & 0 \\ 0 & 2 & 5 & 2 & 0 \\ 0 & 0 & 2 & 5 & 2 \\ 0 & 0 & 0 & 2 & 5 \\ \end{pmatrix}$$

$A$ is symmetric, so it admits a Cholesky decomposition iff it is positive definite. The only methods I know for checking this are:

1. $X^tAX > 0, \quad \forall X \in \mathbb{K}^n- \{0\}$.
2. If $\lambda$ is an eigenvalue of $A$, then $\lambda>0.$

I have failed to prove it using 1 and 2 is taking me so much time. Is there any easier way to do this, given that $A$ is tridiagonal?

• I would think diagonal dominance or Gershgorin discs would be easiest. Looking at the answers it seems so. – mathreadler May 12 '18 at 15:11
• I agree, and I upvoted it but the theorem and its proof are a little bit too much for my linear algebra background. Maybe I should have added "using elementary linear algebra techniques" to the title. – Yagger May 12 '18 at 15:25
• Ah I see now. It was not meant as a critique to the choice of answer. The one you chose was also nice. – mathreadler May 12 '18 at 15:28

Notice $A$ can be rewritten as a sum of 5 matrices. $$A = \left[\begin{smallmatrix} 2 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 3 \end{smallmatrix}\right] + \left[\begin{smallmatrix} 2 & 2 & 0 & 0 & 0\\ 2 & 2 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{smallmatrix}\right] + \left[\begin{smallmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 2 & 2 & 0 & 0\\ 0 & 2 & 2 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{smallmatrix}\right] + \left[\begin{smallmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 2 & 2 & 0\\ 0 & 0 & 2 & 2 & 0\\ 0 & 0 & 0 & 0 & 0 \end{smallmatrix}\right] + \left[\begin{smallmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 2 & 2\\ 0 & 0 & 0 & 2 & 2 \end{smallmatrix}\right]$$ The first matrix is diagonal with positive entries on diagonals, so it is positive definite. The remaining four matrices are clearly positive semi-definite. Being a sum of a positive definite matrix with a bunch of positive semi-definite matrices, $A$ itself is positive definite.

From Gershgorin’s disc theorem and the matrix being symmetric (real eigenvalues) you get that the eigenvalues are in the interval [1,9] and thus positive. Apparently it’s commonly known as Gershgorin’s circle theorem.

This particular matrix is symmetric diagonally dominant (SDD), meaning that the absolute values of each row's off-diagonal entries do not exceed the absolute value of the diagonal, ie. $$\sum_{\substack{j\in[1,n] \\ i \neq j}} \lvert a_{i,j} \rvert \leq \lvert a_{ii} \rvert$$ Since the diagonals are positive, it is positive semidefinite, but it is actually positive definite because the inequality above is strict. This can be seen by splitting $A$ into $A = A' + \varepsilon I$ such that $A'$ remains SDD (and therefore PSD) and $\varepsilon > 0$. Multiplying on both sides by the same vector is always positive, since $\varepsilon I$ is positive definite.

Another approach specific to symmetric tridiagonal matrices is to use the following recurrence. Let $a_i$ be the entries of the main diagonal, and let $b_j$ be the entries of the off-diagonal. Then the determinant is given by \begin{align*} d_{-1} &= 0 \\ d_0 &= 1 \\ d_n &= a_n d_{n-1} - b_{n-1}^2 d_{n-2} \end{align*} This gives the series of determinants of its leading principal minors. In your case, this is: \begin{align*} d_1 = 4\cdot 1 - 0^2 \cdot 0 = 4 \\ d_2 = 5\cdot 4 - 2^2 \cdot 1 = 16 \\ d_3 = 5\cdot 16 - 2^2 \cdot 1 = 76 \\ d_4 = 5\cdot 76- 2^2 \cdot 1 = 376 \\ d_5 = 5\cdot 376- 2^2 \cdot 1 = 1876 \\ \end{align*} Since they're all positive, Sylvester's criterion shows that the matrix is positive-definite.

Since you asked for a "fast" way, I'll note that this takes linear time to execute on a computer. I'll also note that a faster way of doing this on paper (fewer operations) is to divide the recurrence by the previous term to get $s_n = a_n - b^2_{n-1} / s_{n-1}$ with $s_0=1$ and $b_0=0$. If $a_1$ is positive and $s_n$ remains positive throughout, then the recurrence never changes sign, so $d_n$ is positive as before.

It is well-known that the eigenvalues of
$$S=\begin{bmatrix} 0&1&0&0&0\\ 1&0&1&0&0\\ 0&1&0&1&0\\ 0&0&1&0&1\\ 0&0&0&1&0 \end{bmatrix}$$ are $$2\cos\frac{k\pi}{n+1},\ \ \ k=1,\ldots,n.$$ So the eigenvalues of $4I+2S$ are $$4+4\cos\frac{k\pi}{6}>0,\ \ \ k=1,\ldots,5.$$ Being symmetric, this tells us that $$4I+4S=\begin{bmatrix} 4&2&0&0&0\\ 2&4&2&0&0\\ 0&2&4&2&0\\ 0&0&2&4&2\\ 0&0&0&2&4 \end{bmatrix}$$ is positive definite. Now $A$ is also positive definite, as $A=B+4I+2S$, where $$B=\begin{bmatrix} 0&0&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\end{bmatrix}$$ is positive semi definite.

This illustrates Sylvester's Law of Inertia: your matrix is, indeed, positive definite. On the other hand, you actually get a free decomposition this way, so somebody was very careful about setting this one up. With the names I use below $(2Q)^T (2Q) = H$

https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia

$$P^T H P = D$$ $$\left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ - \frac{ 1 }{ 2 } & 1 & 0 & 0 & 0 \\ \frac{ 1 }{ 4 } & - \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ - \frac{ 1 }{ 8 } & \frac{ 1 }{ 4 } & - \frac{ 1 }{ 2 } & 1 & 0 \\ \frac{ 1 }{ 16 } & - \frac{ 1 }{ 8 } & \frac{ 1 }{ 4 } & - \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrrr} 4 & 2 & 0 & 0 & 0 \\ 2 & 5 & 2 & 0 & 0 \\ 0 & 2 & 5 & 2 & 0 \\ 0 & 0 & 2 & 5 & 2 \\ 0 & 0 & 0 & 2 & 5 \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & - \frac{ 1 }{ 2 } & \frac{ 1 }{ 4 } & - \frac{ 1 }{ 8 } & \frac{ 1 }{ 16 } \\ 0 & 1 & - \frac{ 1 }{ 2 } & \frac{ 1 }{ 4 } & - \frac{ 1 }{ 8 } \\ 0 & 0 & 1 & - \frac{ 1 }{ 2 } & \frac{ 1 }{ 4 } \\ 0 & 0 & 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrr} 4 & 0 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 0 & 4 \\ \end{array} \right)$$ $$Q^T D Q = H$$ $$\left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 & 0 & 0 \\ 0 & \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ 0 & 0 & \frac{ 1 }{ 2 } & 1 & 0 \\ 0 & 0 & 0 & \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrrr} 4 & 0 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 0 & 4 \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & \frac{ 1 }{ 2 } & 0 & 0 & 0 \\ 0 & 1 & \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrr} 4 & 2 & 0 & 0 & 0 \\ 2 & 5 & 2 & 0 & 0 \\ 0 & 2 & 5 & 2 & 0 \\ 0 & 0 & 2 & 5 & 2 \\ 0 & 0 & 0 & 2 & 5 \\ \end{array} \right)$$

I have failed to prove it using 1

Note: $$\begin{pmatrix}a&b&c&d&e\end{pmatrix}\begin{pmatrix} 4 & 2 & 0 & 0 & 0 \\ 2 & 5 & 2 & 0 & 0 \\ 0 & 2 & 5 & 2 & 0 \\ 0 & 0 & 2 & 5 & 2 \\ 0 & 0 & 0 & 2 & 5 \\ \end{pmatrix}\begin{pmatrix}a\\b\\c\\d\\e\end{pmatrix}=\\ \begin{pmatrix}4a+2b&2a+5b+2c&2b+5c+2d&2c+5d+2t&2d+5e\end{pmatrix}\begin{pmatrix}a\\b\\c\\d\\e\end{pmatrix}=\\ 4a^2+4ab+5b^2+4bc+5c^2+4cd+5d^2+4de+5e^2=\\ (2a+b)^2+(2b+c)^2+(2c+d)^2+(2d+e)^2+4e^2.$$

Let $I_n$ be $n\times n$ identity matrix and $$J_n=\begin{bmatrix}0 & I_{n-1}\\0 & 0\end{bmatrix}$$ be $n\times n$ Jordan block (with zero blocks being of suitable sizes). Then $$J_n^TJ_n=\begin{bmatrix}0 & 0\\I_{n-1} & 0\end{bmatrix}\begin{bmatrix}0 & I_{n-1}\\0 & 0\end{bmatrix}=\begin{bmatrix}0 & 0\\0 & I_{n-1}\end{bmatrix}$$ and $$(2I_n+J_n)^T(2I_n+J_n)=4I_n+2J_n+2J_n^T+J_n^TJ_n=A$$ is the Cholesky factorization of $A$.

Let use Sylvester's criterion which states that a symmetric matrix is positive-definite if and only if all leading principal minors are positive.